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# Equations and inequalities

## Strategy for Solving Equations

This chapter is all about solving different types of equations for one or two variables. In general, we want to get the unknown variable alone on the left hand side of the equation with all the constants on the right hand side of the equation. For example, in the equation x-1=0x-1=0


, we want to be able to write the equation as x=1x=1.

As we saw in rearranging equations, an equation is like a set of weighing scales that must always be balanced. When we solve equations, we need to keep in mind that what is done to one side must be done to the other.

### Method: Rearranging Equations

You can add, subtract, multiply or divide both sides of an equation by any number you want, as long as you always do it to both sides.

For example, in the equation x+5-1=-6x+5-1=-6


, we want to get xx alone on the left hand side of the equation. This means we need to subtract 5 and add 1 on the left hand side. However, because we need to keep the equation balanced, we also need to subtract 5 and add 1 on the right hand side.

x + 5 - 1 = - 6 x + 5 - 5 - 1 + 1 = - 6 - 5 + 1 x + 0 + 0 = - 11 + 1 x = - 10 x + 5 - 1 = - 6 x + 5 - 5 - 1 + 1 = - 6 - 5 + 1 x + 0 + 0 = - 11 + 1 x = - 10
(1)

In another example, 23x=823x=8, we must divide by 2 and multiply by 3 on the left hand side in order to get xx alone. However, in order to keep the equation balanced, we must also divide by 2 and multiply by 3 on the right hand side.

2 3 x = 8 2 3 x ÷ 2 × 3 = 8 ÷ 2 × 3 2 2 × 3 3 × x = 8 × 3 2 1 × 1 × x = 12 x = 12 2 3 x = 8 2 3 x ÷ 2 × 3 = 8 ÷ 2 × 3 2 2 × 3 3 × x = 8 × 3 2 1 × 1 × x = 12 x = 12
(2)

These are the basic rules to apply when simplifying any equation. In most cases, these rules have to be applied more than once, before we have the unknown variable on the left hand side of the equation.

#### Tip:

The following must also be kept in mind:
1. Division by 0 is undefined.
2. If xy=0xy=0, then x=0x=0 and y0y0, because division by 0 is undefined.

We are now ready to solve some equations!

#### Investigation : 4 = 3 ??

In the following, identify what is wrong.

x = 2 4 x - 8 = 3 x - 6 4 ( x - 2 ) = 3 ( x - 2 ) 4 ( x - 2 ) ( x - 2 ) = 3 ( x - 2 ) ( x - 2 ) 4 = 3 x = 2 4 x - 8 = 3 x - 6 4 ( x - 2 ) = 3 ( x - 2 ) 4 ( x - 2 ) ( x - 2 ) = 3 ( x - 2 ) ( x - 2 ) 4 = 3
(3)

## Solving Linear Equations

The simplest equation to solve is a linear equation. A linear equation is an equation where the power of the variable(letter, e.g. xx) is 1(one). The following are examples of linear equations.

2 x + 2 = 1 2 - x 3 x + 1 = 2 4 3 x - 6 = 7 x + 2 2 x + 2 = 1 2 - x 3 x + 1 = 2 4 3 x - 6 = 7 x + 2
(4)

In this section, we will learn how to find the value of the variable that makes both sides of the linear equation true. For example, what value of xx makes both sides of the very simple equation, x+1=1x+1=1 true.

Since the definition of a linear equation is that if the variable has a highest power of one (1), there is at most one solution or root for the equation.

This section relies on all the methods we have already discussed: multiplying out expressions, grouping terms and factorisation. Make sure that you are comfortable with these methods, before trying out the work in the rest of this chapter.

2 x + 2 = 1 2 x = 1 - 2 ( like terms together ) 2 x = - 1 ( simplified as much as possible ) 2 x + 2 = 1 2 x = 1 - 2 ( like terms together ) 2 x = - 1 ( simplified as much as possible )
(5)

Now we see that 2x=-12x=-1. This means if we divide both sides by 2, we will get:

x = - 1 2 x = - 1 2
(6)

If we substitute x=-12x=-12, back into the original equation, we get:

LHS = 2 x + 2 = 2 ( - 1 2 ) + 2 = - 1 + 2 = 1 and RHS = 1 LHS = 2 x + 2 = 2 ( - 1 2 ) + 2 = - 1 + 2 = 1 and RHS = 1
(7)

That is all that there is to solving linear equations.

### Tip: Solving Equations:

When you have found the solution to an equation, substitute the solution into the original equation, to check your answer.

### Method: Solving Linear Equations

The general steps to solve linear equations are:

1. Expand brackets: Expand (Remove) all brackets that are in the equation.
2. Rearrange: "Move" all terms with the variable to the left hand side of the equation, and all constant terms (the numbers) to the right hand side of the equals sign. Bearing in mind that the sign of the terms will change from (++) to (--) or vice versa, as they "cross over" the equals sign.
3. Group like terms: Group all like terms together and simplify as much as possible.
4. Factorise: If necessary factorise.
5. Write solution: Find the solution and write down the answer(s).
6. Check: Substitute solution into original equation to check answer.

Figure 1
Khan academy video on equations - 1

#### Exercise 1: Solving Linear Equations

Solve for xx: 4-x=44-x=4

##### Solution
1. Step 1. Determine what is given and what is required :

We are given 4-x=44-x=4 and are required to solve for xx.

2. Step 2. Determine how to approach the problem :

Since there are no brackets, we can start with rearranging and then grouping like terms.

3. Step 3. Solve the problem :
4 - x = 4 - x = 4 - 4 ( Rearrange ) - x = 0 ( group like terms ) x = 0 4 - x = 4 - x = 4 - 4 ( Rearrange ) - x = 0 ( group like terms ) x = 0
(8)
4. Step 4. Check the answer :

Substitute solution into original equation:

4 - 0 = 4 4 = 4 4 - 0 = 4 4 = 4
(9)

Since both sides are equal, the answer is correct.

5. Step 5. Write the final answer :

The solution of 4-x=44-x=4 is x=0x=0.

#### Exercise 2: Solving Linear Equations

Solve for xx: 4(2x-9)-4x=4-6x4(2x-9)-4x=4-6x

##### Solution
1. Step 1. Determine what is given and what is required :

We are given 4(2x-9)-4x=4-6x4(2x-9)-4x=4-6x and are required to solve for xx.

2. Step 2. Determine how to approach the problem :

We start with expanding the brackets, then rearranging, then grouping like terms and then simplifying.

3. Step 3. Solve the problem :
4 ( 2 x - 9 ) - 4 x = 4 - 6 x 8 x - 36 - 4 x = 4 - 6 x ( expand the brackets ) 8 x - 4 x + 6 x = 4 + 36 ( Rearrange ) ( 8 x - 4 x + 6 x ) = ( 4 + 36 ) ( group like terms ) 10 x = 40 ( simplify grouped terms ) 10 10 x = 40 10 ( divide both sides by 10 ) x = 4 4 ( 2 x - 9 ) - 4 x = 4 - 6 x 8 x - 36 - 4 x = 4 - 6 x ( expand the brackets ) 8 x - 4 x + 6 x = 4 + 36 ( Rearrange ) ( 8 x - 4 x + 6 x ) = ( 4 + 36 ) ( group like terms ) 10 x = 40 ( simplify grouped terms ) 10 10 x = 40 10 ( divide both sides by 10 ) x = 4
(10)
4. Step 4. Check the answer :

Substitute solution into original equation:

4 ( 2 ( 4 ) - 9 ) - 4 ( 4 ) = 4 - 6 ( 4 ) 4 ( 8 - 9 ) - 16 = 4 - 24 4 ( - 1 ) - 16 = - 20 - 4 - 16 = - 20 - 20 = - 20 4 ( 2 ( 4 ) - 9 ) - 4 ( 4 ) = 4 - 6 ( 4 ) 4 ( 8 - 9 ) - 16 = 4 - 24 4 ( - 1 ) - 16 = - 20 - 4 - 16 = - 20 - 20 = - 20
(11)

Since both sides are equal to -20-20, the answer is correct.

5. Step 5. Write the final answer :

The solution of 4(2x-9)-4x=4-6x4(2x-9)-4x=4-6x is x=4x=4.

#### Exercise 3: Solving Linear Equations

Solve for xx: 2-x3x+1=22-x3x+1=2

##### Solution
1. Step 1. Determine what is given and what is required :

We are given 2-x3x+1=22-x3x+1=2 and are required to solve for xx.

2. Step 2. Determine how to approach the problem :

Since there is a denominator of (3x+13x+1), we can start by multiplying both sides of the equation by (3x+13x+1). But because division by 0 is not permissible, there is a restriction on a value for x. (x-13x-13)

3. Step 3. Solve the problem :
2 - x 3 x + 1 = 2 ( 2 - x ) = 2 ( 3 x + 1 ) 2 - x = 6 x + 2 ( expand brackets ) - x - 6 x = 2 - 2 ( rearrange) - 7 x = 0 ( simplify grouped terms ) x = 0 ÷ ( - 7 ) x = 0 ( zero divided by any number is 0) 2 - x 3 x + 1 = 2 ( 2 - x ) = 2 ( 3 x + 1 ) 2 - x = 6 x + 2 ( expand brackets ) - x - 6 x = 2 - 2 ( rearrange) - 7 x = 0 ( simplify grouped terms ) x = 0 ÷ ( - 7 ) x = 0 ( zero divided by any number is 0)
(12)
4. Step 4. Check the answer :

Substitute solution into original equation:

2 - ( 0 ) 3 ( 0 ) + 1 = 2 2 1 = 2 2 - ( 0 ) 3 ( 0 ) + 1 = 2 2 1 = 2
(13)

Since both sides are equal to 2, the answer is correct.

5. Step 5. Write the final answer :

The solution of 2-x3x+1=22-x3x+1=2 is x=0x=0.

#### Exercise 4: Solving Linear Equations

Solve for xx: 43x-6=7x+243x-6=7x+2

##### Solution
1. Step 1. Determine what is given and what is required :

We are given 43x-6=7x+243x-6=7x+2 and are required to solve for xx.

2. Step 2. Determine how to approach the problem :

We start with multiplying each of the terms in the equation by 3, then grouping like terms and then simplifying.

3. Step 3. Solve the problem :
4 3 x - 6 = 7 x + 2 4 x - 18 = 21 x + 6 ( each term is multiplied by 3 ) 4 x - 21 x = 6 + 18 ( rearrange) - 17 x = 24 ( simplify grouped terms ) - 17 - 17 x = 24 - 17 ( divide both sides by - 17 ) x = - 24 17 4 3 x - 6 = 7 x + 2 4 x - 18 = 21 x + 6 ( each term is multiplied by 3 ) 4 x - 21 x = 6 + 18 ( rearrange) - 17 x = 24 ( simplify grouped terms ) - 17 - 17 x = 24 - 17 ( divide both sides by - 17 ) x = - 24 17
(14)
4. Step 4. Check the answer :

Substitute solution into original equation:

4 3 × - 24 17 - 6 = 7 × - 24 17 + 2 4 × ( - 8 ) ( 17 ) - 6 = 7 × ( - 24 ) 17 + 2 ( - 32 ) 17 - 6 = - 168 17 + 2 - 32 - 102 17 = ( - 168 ) + 34 17 - 134 17 = - 134 17 4 3 × - 24 17 - 6 = 7 × - 24 17 + 2 4 × ( - 8 ) ( 17 ) - 6 = 7 × ( - 24 ) 17 + 2 ( - 32 ) 17 - 6 = - 168 17 + 2 - 32 - 102 17 = ( - 168 ) + 34 17 - 134 17 = - 134 17
(15)

Since both sides are equal to -13417-13417, the answer is correct.

5. Step 5. Write the final answer :

The solution of 43x-6=7x+243x-6=7x+2 is,   x=-2417x=-2417.

#### Solving Linear Equations

1. Solve for yy: 2y-3=72y-3=7

2. Solve for yy: -3y=0-3y=0

3. Solve for yy: 4y=164y=16

4. Solve for yy: 12y+0=14412y+0=144

5. Solve for yy: 7+5y=627+5y=62

6. Solve for xx: 55=5x+3455=5x+34

7. Solve for xx: 5x=3x+455x=3x+45

8. Solve for xx: 23x-12=6+2x23x-12=6+2x

9. Solve for xx: 12-6x+34x=2x-24-6412-6x+34x=2x-24-64

10. Solve for xx: 6x+3x=4-5(2x-3)6x+3x=4-5(2x-3)

11. Solve for pp: 18-2p=p+918-2p=p+9

12. Solve for pp: 4p=16244p=1624

13. Solve for pp: 41=p241=p2

14. Solve for pp: -(-16-p)=13p-1-(-16-p)=13p-1

15. Solve for pp: 6p-2+2p=-2+4p+86p-2+2p=-2+4p+8

16. Solve for ff: 3f-10=103f-10=10

17. Solve for ff: 3f+16=4f-103f+16=4f-10

18. Solve for ff: 10f+5+0=-2f+-3f+8010f+5+0=-2f+-3f+80

19. Solve for ff: 8(f-4)=5(f-4)8(f-4)=5(f-4)

20. Solve for ff: 6=6(f+7)+5f6=6(f+7)+5f


A quadratic equation is an equation where the power of the variable is at most 2. The following are examples of quadratic equations.

2 x 2 + 2 x = 1 2 - x 3 x + 1 = 2 x 4 3 x - 6 = 7 x 2 + 2 2 x 2 + 2 x = 1 2 - x 3 x + 1 = 2 x 4 3 x - 6 = 7 x 2 + 2
(16)

Quadratic equations differ from linear equations by the fact that a linear equation only has one solution, while a quadratic equation has at most two solutions. However, there are some special situations when a quadratic equation only has one solution.

We solve quadratic equations by factorisation, that is writing the quadratic as a product of two expressions in brackets. For example, we know that:

( x + 1 ) ( 2 x - 3 ) = 2 x 2 - x - 3 . ( x + 1 ) ( 2 x - 3 ) = 2 x 2 - x - 3 .
(17)

In order to solve:

2 x 2 - x - 3 = 0 2 x 2 - x - 3 = 0
(18)

we need to be able to write 2x2-x-32x2-x-3 as (x+1)(2x-3)(x+1)(2x-3), which we already know how to do. The reason for equating to zero and factoring is that if we attempt to solve it in a 'normal' way, we may miss one of the solutions. On the other hand, if we have the (non-linear) equation f(x)g(x)=0f(x)g(x)=0, for some functions ff and gg, we know that the solution is f(x)=0f(x)=0 OR g(x)=0g(x)=0, which allows us to find BOTH solutions (or know that there is only one solution if it turns out that f=gf=g).

### Investigation : Factorising a Quadratic

1. x + x 2 x + x 2
2. x 2 + 1 + 2 x x 2 + 1 + 2 x
3. x 2 - 4 x + 5 x 2 - 4 x + 5
4. 16 x 2 - 9 16 x 2 - 9
5. 4 x 2 + 4 x + 1 4 x 2 + 4 x + 1

Being able to factorise a quadratic means that you are one step away from solving a quadratic equation. For example, x2-3x-2=0x2-3x-2=0 can be written as (x-1)(x-2)=0(x-1)(x-2)=0. This means that both x-1=0x-1=0 and x-2=0x-2=0, which gives x=1x=1 and x=2x=2 as the two solutions to the quadratic equation x2-3x-2=0x2-3x-2=0.

1. First divide the entire equation by any common factor of the coefficients, so as to obtain an equation of the form ax2+bx+c=0ax2+bx+c=0 where aa, bb and cc have no common factors. For example, 2x2+4x+2=02x2+4x+2=0

can be written as x2+2x+1=0x2+2x+1=0

by dividing by 2.
2. Write ax2+bx+cax2+bx+c in terms of its factors (rx+s)(ux+v)(rx+s)(ux+v). This means (rx+s)(ux+v)=0(rx+s)(ux+v)=0.
3. Once writing the equation in the form (rx+s)(ux+v)=0(rx+s)(ux+v)=0, it then follows that the two solutions are x=-srx=-sr or x=-uvx=-uv.
4. For each solution substitute the value into the original equation to check whether it is valid

There are two solutions to a quadratic equation, because any one of the values can solve the equation.

Figure 2
Khan academy video on equations - 3

#### Exercise 5: Solving Quadratic Equations

Solve for xx: 3x2+2x-1=03x2+2x-1=0

##### Solution
1. Step 1. Find the factors of 3x2+2x-13x2+2x-1 :

As we have seen the factors of 3x2+2x-13x2+2x-1 are (x+1)(x+1) and (3x-1)(3x-1).

2. Step 2. Write the equation with the factors :
( x + 1 ) ( 3 x - 1 ) = 0 ( x + 1 ) ( 3 x - 1 ) = 0
(19)
3. Step 3. Determine the two solutions :

We have

x + 1 = 0 x + 1 = 0
(20)

or

3 x - 1 = 0 3 x - 1 = 0
(21)

Therefore, x=-1x=-1 or x=13x=13.

4. Step 4. Check the solutions: We substitute the answers back into the original equation and for both answers we find that the equation is true.
5. Step 5. Write the final answer :

3x2+2x-1=03x2+2x-1=0


for x=-1x=-1 or x=13x=13.

Sometimes an equation might not look like a quadratic at first glance but turns into one with a simple operation or two. Remember that you have to do the same operation on both sides of the equation for it to remain true.

You might need to do one (or a combination) of:

• Multiply both sides: For example,
ax+b=cxx(ax+b)=x(cx)ax2+bx=cax+b=cxx(ax+b)=x(cx)ax2+bx=c
(22)
• Invert both sides: This is raising both sides to the power of -1-1. For example,
1ax2+bx=c(1ax2+bx)-1=(c)-1ax2+bx1=1cax2+bx=1c1ax2+bx=c(1ax2+bx)-1=(c)-1ax2+bx1=1cax2+bx=1c
(23)
• Square both sides: This is raising both sides to the power of 2. For example,
ax2+bx=c(ax2+bx)2=c2ax2+bx=c2ax2+bx=c(ax2+bx)2=c2ax2+bx=c2
(24)

You can combine these in many ways and so the best way to develop your intuition for the best thing to do is practice problems. A combined set of operations could be, for example,

1 a x 2 + b x = c ( 1 a x 2 + b x ) - 1 = ( c ) - 1 ( invert both sides ) a x 2 + b x 1 = 1 c a x 2 + b x = 1 c ( a x 2 + b x ) 2 = ( 1 c ) 2 ( square both sides ) a x 2 + b x = 1 c 2 1 a x 2 + b x = c ( 1 a x 2 + b x ) - 1 = ( c ) - 1 ( invert both sides ) a x 2 + b x 1 = 1 c a x 2 + b x = 1 c ( a x 2 + b x ) 2 = ( 1 c ) 2 ( square both sides ) a x 2 + b x = 1 c 2
(25)

#### Exercise 6: Solving Quadratic Equations

Solve for xx: x+2=xx+2=x

##### Solution
1. Step 1. Square both sides of the equation :

Both sides of the equation should be squared to remove the square root sign.

x + 2 = x 2 x + 2 = x 2
(26)
2. Step 2. Write equation in the form ax2+bx+c=0ax2+bx+c=0 :
x + 2 = x 2 ( subtract x 2 from both sides ) x + 2 - x 2 = 0 ( divide both sides by - 1 ) - x - 2 + x 2 = 0 x 2 - x + 2 = 0 x + 2 = x 2 ( subtract x 2 from both sides ) x + 2 - x 2 = 0 ( divide both sides by - 1 ) - x - 2 + x 2 = 0 x 2 - x + 2 = 0
(27)
3. Step 3. Factorise the quadratic :
x 2 - x + 2 x 2 - x + 2
(28)

The factors of x2-x+2x2-x+2 are (x-2)(x+1)(x-2)(x+1).

4. Step 4. Write the equation with the factors :
( x - 2 ) ( x + 1 ) = 0 ( x - 2 ) ( x + 1 ) = 0
(29)
5. Step 5. Determine the two solutions :

We have

x + 1 = 0 x + 1 = 0
(30)

or

x - 2 = 0 x - 2 = 0
(31)

Therefore, x=-1x=-1 or x=2x=2.

6. Step 6. Check whether solutions are valid :

Substitute x=-1x=-1


into the original equation x+2=xx+2=x:

LHS = ( - 1 ) + 2 = 1 = 1 but RHS = ( - 1 ) LHS = ( - 1 ) + 2 = 1 = 1 but RHS = ( - 1 )
(32)

Therefore LHS RHS. The sides of an equation must always balance, a potential solution that does not balance the equation is not valid. In this case the equation does not balance.

Therefore x-1x-1.

Now substitute x=2x=2 into original equation x+2=xx+2=x:

LHS = 2 + 2 = 4 = 2 and RHS = 2 LHS = 2 + 2 = 4 = 2 and RHS = 2
(33)

Therefore LHS = RHS

Therefore x=2x=2 is the only valid solution

7. Step 7. Write the final answer :

x+2=xx+2=x for x=2x=2 only.

#### Exercise 7: Solving Quadratic Equations

Solve the equation: x2+3x-4=0x2+3x-4=0.

##### Solution
1. Step 1. Check if the equation is in the form ax2+bx+c=0ax2+bx+c=0 :

The equation is in the required form, with a=1a=1.

2. Step 2. Factorise the quadratic :

You need the factors of 1 and 4 so that the middle term is +3+3 So the factors are:

( x - 1 ) ( x + 4 ) ( x - 1 ) ( x + 4 )

3. Step 3. Solve the quadratic equation :
x 2 + 3 x - 4 = ( x - 1 ) ( x + 4 ) = 0 x 2 + 3 x - 4 = ( x - 1 ) ( x + 4 ) = 0
(34)

Therefore x=1x=1 or x=-4x=-4.

4. Step 4. Check the solutions:
12+3(1)-4=012+3(1)-4=0
(35)
(-4)2+3(-4)-4=0(-4)2+3(-4)-4=0
(36)

Both solutions are valid.

5. Step 5. Write the final solution :

Therefore the solutions are x=1x=1 or x=-4x=-4.

#### Exercise 8: Solving Quadratic Equations

Find the roots of the quadratic equation 0=-2x2+4x-20=-2x2+4x-2.

##### Solution
1. Step 1. Determine whether the equation is in the form ax2+bx+c=0ax2+bx+c=0, with no common factors. :

There is a common factor: -2. Therefore, divide both sides of the equation by -2.

- 2 x 2 + 4 x - 2 = 0 x 2 - 2 x + 1 = 0 - 2 x 2 + 4 x - 2 = 0 x 2 - 2 x + 1 = 0
(37)
2. Step 2. Factorise x2-2x+1x2-2x+1 :

The middle term is negative. Therefore, the factors are (x-1)(x-1)(x-1)(x-1)

If we multiply out (x-1)(x-1)(x-1)(x-1), we get x2-2x+1x2-2x+1.

3. Step 3. Solve the quadratic equation :
x 2 - 2 x + 1 = ( x - 1 ) ( x - 1 ) = 0 x 2 - 2 x + 1 = ( x - 1 ) ( x - 1 ) = 0
(38)

In this case, the quadratic is a perfect square, so there is only one solution for xx: x=1x=1.

4. Step 4. Check the solution:

-2(1)2+4(1)-2=0-2(1)2+4(1)-2=0.

5. Step 5. Write the final solution :

The root of 0=-2x2+4x-20=-2x2+4x-2 is x=1x=1.

1. Solve for xx: (3x+2)(3x-4)=0(3x+2)(3x-4)=0

2. Solve for xx: (5x-9)(x+6)=0(5x-9)(x+6)=0

3. Solve for xx: (2x+3)(2x-3)=0(2x+3)(2x-3)=0

4. Solve for xx: (2x+1)(2x-9)=0(2x+1)(2x-9)=0

5. Solve for xx: (2x-3)(2x-3)=0(2x-3)(2x-3)=0

6. Solve for xx: 20x+25x2=020x+25x2=0

7. Solve for xx: 4x2-17x-77=04x2-17x-77=0

8. Solve for xx: 2x2-5x-12=02x2-5x-12=0

9. Solve for xx: -75x2+290x-240=0-75x2+290x-240=0

10. Solve for xx: 2x=13x2-3x+14232x=13x2-3x+1423

11. Solve for xx: x2-4x=-4x2-4x=-4

12. Solve for xx: -x2+4x-6=4x2-5x+3-x2+4x-6=4x2-5x+3

13. Solve for xx: x2=3xx2=3x

14. Solve for xx: 3x2+10x-25=03x2+10x-25=0

15. Solve for xx: x2-x+3x2-x+3


16. Solve for xx: x2-4x+4=0x2-4x+4=0

17. Solve for xx: x2-6x=7x2-6x=7

18. Solve for xx: 14x2+5x=614x2+5x=6

19. Solve for xx: 2x2-2x=122x2-2x=12

20. Solve for xx: 3x2+2y-6=x2-x+23x2+2y-6=x2-x+2


## Exponential Equations of the Form ka(x+p)=mka(x+p)=m

Exponential equations generally have the unknown variable as the power. The following are examples of exponential equations:

2 x = 1 2 - x 3 x + 1 = 2 4 3 - 6 = 7 x + 2 2 x = 1 2 - x 3 x + 1 = 2 4 3 - 6 = 7 x + 2
(39)

You should already be familiar with exponential notation. Solving exponential equations is simple, if we remember how to apply the laws of exponentials.

### Investigation : Solving Exponential Equations

Solve the following equations by completing the table:

 2 x = 2 2 x = 2 x x -3 -2 -1 0 1 2 3 2 x 2 x
 3 x = 9 3 x = 9 x x -3 -2 -1 0 1 2 3 3 x 3 x
 2 x + 1 = 8 2 x + 1 = 8 x x -3 -2 -1 0 1 2 3 2 x + 1 2 x + 1

### Algebraic Solution

Definition 1: Equality for Exponential Functions

If aa is a positive number such that a>0a>0, (except when a=1a=1


) then:

a x = a y a x = a y
(40)

if and only if:

x = y x = y
(41)
(If a=1a=1, then xx and yy can differ)

This means that if we can write all terms in an equation with the same base, we can solve the exponential equations by equating the indices. For example take the equation 3x+1=93x+1=9. This can be written as:

3 x + 1 = 3 2 . 3 x + 1 = 3 2 .
(42)

Since the bases are equal (to 3), we know that the exponents must also be equal. Therefore we can write:

x + 1 = 2 . x + 1 = 2 .
(43)

This gives:

x = 1 . x = 1 .
(44)

#### Method: Solving Exponential Equations

1. Try to write all terms with the same base.
2. Equate the exponents of the bases of the left and right hand side of the equation.
3. Solve the equation obtained in the previous step.
4. Check the solutions
##### Investigation : Exponential Numbers

Write the following with the same base. The base is the first in the list. For example, in the list 2, 4, 8, the base is two and we can write 4 as 2222.

1. 2,4,8,16,32,64,128,512,1024
2. 3,9,27,81,243
3. 5,25,125,625
4. 13,169
5. 2x2x, 4x24x2, 8x38x3, 49x849x8
##### Exercise 9: Solving Exponential Equations

Solve for xx: 2x=22x=2

###### Solution
1. Step 1. Try to write all terms with the same base. :

All terms are written with the same base.

2 x = 2 1 2 x = 2 1
(45)
2. Step 2. Equate the exponents :
x = 1 x = 1
(46)
LHS = 2 x = 2 1 = 2 RHS = 2 1 = 2 = LHS LHS = 2 x = 2 1 = 2 RHS = 2 1 = 2 = LHS
(47)

Since both sides are equal, the answer is correct.

4. Step 4. Write the final answer :
x = 1 x = 1
(48)

is the solution to 2x=22x=2.

##### Exercise 10: Solving Exponential Equations

Solve:

2 x + 4 = 4 2 x 2 x + 4 = 4 2 x
(49)
###### Solution
1. Step 1. Try to write all terms with the same base. :
2 x + 4 = 4 2 x 2 x + 4 = 2 2 ( 2 x ) 2 x + 4 = 2 4 x 2 x + 4 = 4 2 x 2 x + 4 = 2 2 ( 2 x ) 2 x + 4 = 2 4 x
(50)
2. Step 2. Equate the exponents :
x + 4 = 4 x x + 4 = 4 x
(51)
3. Step 3. Solve for xx :
x + 4 = 4 x x - 4 x = - 4 - 3 x = - 4 x = - 4 - 3 x = 4 3 x + 4 = 4 x x - 4 x = - 4 - 3 x = - 4 x = - 4 - 3 x = 4 3
(52)
LHS = 2 x + 4 = 2 ( 4 3 + 4 ) = 2 16 3 = ( 2 16 ) 1 3 RHS = 4 2 x = 4 2 ( 4 3 ) = 4 8 3 = ( 4 8 ) 1 3 = ( ( 2 2 ) 8 ) 1 3 = ( 2 16 ) 1 3 = LHS LHS = 2 x + 4 = 2 ( 4 3 + 4 ) = 2 16 3 = ( 2 16 ) 1 3 RHS = 4 2 x = 4 2 ( 4 3 ) = 4 8 3 = ( 4 8 ) 1 3 = ( ( 2 2 ) 8 ) 1 3 = ( 2 16 ) 1 3 = LHS
(53)

Since both sides are equal, the answer is correct.

5. Step 5. Write the final answer :
x = 4 3 x = 4 3
(54)

is the solution to 2x+4=42x2x+4=42x.

##### Solving Exponential Equations
1. Solve the following exponential equations:
1. 2x+5=252x+5=25
2. 32x+1=3332x+1=33
3. 52x+2=5352x+2=53
4. 65-x=61265-x=612
5. 64x+1=162x+564x+1=162x+5
6. 125x=5125x=5
2. Solve: 39x-2=2739x-2=27
3. Solve for kk: 81k+2=27k+481k+2=27k+4
4. The growth of an algae in a pond can be modeled by the function f(t)=2tf(t)=2t. Find the value of tt such that f(t)=128f(t)=128

5. Solve for xx: 25(1-2x)=5425(1-2x)=54

6. Solve for xx: 27x×9x-2=127x×9x-2=1


## Linear Inequalities

### Investigation : Inequalities on a Number Line

Represent the following on number lines:

1. x = 4 x = 4
2. x < 4 x < 4
3. x 4 x 4
4. x 4 x 4
5. x > 4 x > 4

A linear inequality is similar to a linear equation in that the largest exponent of a variable is 1. The following are examples of linear inequalities.

2 x + 2 1 2 - x 3 x + 1 2 4 3 x - 6 < 7 x + 2 2 x + 2 1 2 - x 3 x + 1 2 4 3 x - 6 < 7 x + 2
(55)

The methods used to solve linear inequalities are identical to those used to solve linear equations. The only difference occurs when there is a multiplication or a division that involves a minus sign. For example, we know that 8>68>6. If both sides of the inequality are divided by -2-2, -4-4 is not greater than -3-3. Therefore, the inequality must switch around, making -4<-3-4<-3.

### Tip:

When you divide or multiply both sides of an inequality by any number with a minus sign, the direction of the inequality changes. For this reason you cannot divide or multiply by a variable.

For example, if x<1x<1, then -x>-1-x>-1.

In order to compare an inequality to a normal equation, we shall solve an equation first. Solve 2x+2=12x+2=1.

2 x + 2 = 1 2 x = 1 - 2 2 x = - 1 x = - 1 2 2 x + 2 = 1 2 x = 1 - 2 2 x = - 1 x = - 1 2
(56)

If we represent this answer on a number line, we get

Now let us solve the inequality 2x+212x+21.

2 x + 2 1 2 x 1 - 2 2 x - 1 x - 1 2 2 x + 2 1 2 x 1 - 2 2 x - 1 x - 1 2
(57)

If we represent this answer on a number line, we get

As you can see, for the equation, there is only a single value of xx for which the equation is true. However, for the inequality, there is a range of values for which the inequality is true. This is the main difference between an equation and an inequality.

Figure 5
Khan academy video on inequalities - 1

Figure 6
Khan academy video on inequalities - 2

### Exercise 11: Linear Inequalities

Solve for rr: 6-r>26-r>2

#### Solution

1. Step 1. Move all constants to the RHS :
- r > 2 - 6 - r > - 4 - r > 2 - 6 - r > - 4
(58)
2. Step 2. Multiply both sides by -1 :

When you multiply by a minus sign, the direction of the inequality changes.

r < 4 r < 4
(59)
3. Step 3. Represent answer graphically :

### Exercise 12: Linear Inequalities

Solve for qq: 4q+3<2(q+3)4q+3<2(q+3) and represent the solution on a number line.

#### Solution

1. Step 1. Expand all brackets :
4 q + 3 < 2 ( q + 3 ) 4 q + 3 < 2 q + 6 4 q + 3 < 2 ( q + 3 ) 4 q + 3 < 2 q + 6
(60)
2. Step 2. Move all constants to the RHS and all unknowns to the LHS :
4 q + 3 < 2 q + 6 4 q - 2 q < 6 - 3 2 q < 3 4 q + 3 < 2 q + 6 4 q - 2 q < 6 - 3 2 q < 3
(61)
3. Step 3. Solve inequality :
2 q < 3 Divide both sides by 2 q < 3 2 2 q < 3 Divide both sides by 2 q < 3 2
(62)
4. Step 4. Represent answer graphically :

### Exercise 13: Compound Linear Inequalities

Solve for xx: 5x+3<85x+3<8 and represent solution on a number line.

#### Solution

1. Step 1. Subtract 3 from Left, middle and right of inequalities :
5 - 3 x + 3 - 3 < 8 - 3 2 x < 5 5 - 3 x + 3 - 3 < 8 - 3 2 x < 5
(63)
2. Step 2. Represent answer graphically :

### Linear Inequalities

1. Solve for xx and represent the solution graphically:
1. 3x+4>5x+83x+4>5x+8
2. 3(x-1)-26x+43(x-1)-26x+4
3. x-73>2x-32x-73>2x-32
4. -4(x-1)<x+2-4(x-1)<x+2
5. 12x+13(x-1)56x-1312x+13(x-1)56x-13

2. Solve the following inequalities. Illustrate your answer on a number line if xx is a real number.
1. -2x-1<3-2x-1<3
2. -5<2x-37-5<2x-37
3. Solve for xx: 7(3x+2)-5(2x-3)>77(3x+2)-5(2x-3)>7

Illustrate this answer on a number line.


## Linear Simultaneous Equations

Thus far, all equations that have been encountered have one unknown variable that must be solved for. When two unknown variables need to be solved for, two equations are required and these equations are known as simultaneous equations. The solutions to the system of simultaneous equations are the values of the unknown variables which satisfy the system of equations simultaneously, that means at the same time. In general, if there are nn unknown variables, then nn equations are required to obtain a solution for each of the nn variables.

An example of a system of simultaneous equations is:

2 x + 2 y = 1 2 - x 3 y + 1 = 2 2 x + 2 y = 1 2 - x 3 y + 1 = 2
(64)

### Finding solutions

In order to find a numerical value for an unknown variable, one must have at least as many independent equations as variables. We solve simultaneous equations graphically and algebraically.

Figure 10
Khan academy video on simultaneous equations - 1

### Graphical Solution

Simultaneous equations can be solved graphically. If the graph corresponding to each equation is drawn, then the solution to the system of simultaneous equations is the co-ordinate of the point at which both graphs intersect.

x = 2 y y = 2 x - 3 x = 2 y y = 2 x - 3
(65)

Draw the graphs of the two equations in Equation 65.

The intersection of the two graphs is (2,1)(2,1). So the solution to the system of simultaneous equations in Equation 65 is y=1y=1 and x=2x=2.

This can be shown algebraically as:

x = 2 y y = 2 ( 2 y ) - 3 y - 4 y = - 3 - 3 y = - 3 y = 1 Substitute into the first equation: x = 2 ( 1 ) = 2 x = 2 y y = 2 ( 2 y ) - 3 y - 4 y = - 3 - 3 y = - 3 y = 1 Substitute into the first equation: x = 2 ( 1 ) = 2
(66)

#### Exercise 14: Simultaneous Equations

Solve the following system of simultaneous equations graphically.

4 y + 3 x = 100 4 y - 19 x = 12 4 y + 3 x = 100 4 y - 19 x = 12
(67)
##### Solution
1. Step 1. Draw the graphs corresponding to each equation. :

For the first equation:

4 y + 3 x = 100 4 y = 100 - 3 x y = 25 - 3 4 x 4 y + 3 x = 100 4 y = 100 - 3 x y = 25 - 3 4 x
(68)

and for the second equation:

4 y - 19 x = 12 4 y = 19 x + 12 y = 19 4 x + 3 4 y - 19 x = 12 4 y = 19 x + 12 y = 19 4 x + 3
(69)

2. Step 2. Find the intersection of the graphs. :

The graphs intersect at (4,22)(4,22).

3. Step 3. Write the solution of the system of simultaneous equations as given by the intersection of the graphs. :
x = 4 y = 22 x = 4 y = 22
(70)

### Solution by Substitution

A common algebraic technique is the substitution method: try to solve one of the equations for one of the variables and substitute the result into the other equations, thereby reducing the number of equations and the number of variables by 1. Continue until you reach a single equation with a single variable, which (hopefully) can be solved; back substitution then allows checking the values for the other variables.

In the example Equation 64, we first solve the first equation for xx:

x = 1 2 - y x = 1 2 - y
(71)

and substitute this result into the second equation:

2 - x 3 y + 1 = 2 2 - ( 1 2 - y ) 3 y + 1 = 2 2 - ( 1 2 - y ) = 2 ( 3 y + 1 ) 2 - 1 2 + y = 6 y + 2 y - 6 y = - 2 + 1 2 + 2 - 5 y = 1 2 y = - 1 10 2 - x 3 y + 1 = 2 2 - ( 1 2 - y ) 3 y + 1 = 2 2 - ( 1 2 - y ) = 2 ( 3 y + 1 ) 2 - 1 2 + y = 6 y + 2 y - 6 y = - 2 + 1 2 + 2 - 5 y = 1 2 y = - 1 10
(72)
x = 1 2 - y = 1 2 - ( - 1 10 ) = 6 10 = 3 5 x = 1 2 - y = 1 2 - ( - 1 10 ) = 6 10 = 3 5
(73)

The solution for the system of simultaneous equations Equation 64 is:

x = 3 5 y = - 1 10 x = 3 5 y = - 1 10
(74)

#### Exercise 15: Simultaneous Equations

Solve the following system of simultaneous equations:

4 y + 3 x = 100 4 y - 19 x = 12 4 y + 3 x = 100 4 y - 19 x = 12
(75)
##### Solution
1. Step 1. Decide how to solve the problem: If the question does not explicitly ask for a graphical solution, then the system of equations should be solved algebraically.
2. Step 2. Make xx the subject of the first equation. :
4 y + 3 x = 100 3 x = 100 - 4 y x = 100 - 4 y 3 4 y + 3 x = 100 3 x = 100 - 4 y x = 100 - 4 y 3
(76)
3. Step 3. Substitute the value obtained for xx into the second equation. :
4 y - 19 ( 100 - 4 y 3 ) = 12 12 y - 19 ( 100 - 4 y ) = 36 12 y - 1900 + 76 y = 36 88 y = 1936 y = 22 4 y - 19 ( 100 - 4 y 3 ) = 12 12 y - 19 ( 100 - 4 y ) = 36 12 y - 1900 + 76 y = 36 88 y = 1936 y = 22
(77)
4. Step 4. Substitute into the equation for xx. :
x = 100 - 4 ( 22 ) 3 = 100 - 88 3 = 12 3 = 4 x = 100 - 4 ( 22 ) 3 = 100 - 88 3 = 12 3 = 4
(78)
5. Step 5. Substitute the values for xx and yy into both equations to check the solution. :
4 ( 22 ) + 3 ( 4 ) = 88 + 12 = 100 4 ( 22 ) - 19 ( 4 ) = 88 - 76 = 12 4 ( 22 ) + 3 ( 4 ) = 88 + 12 = 100 4 ( 22 ) - 19 ( 4 ) = 88 - 76 = 12
(79)

#### Exercise 16: Bicycles and Tricycles

A shop sells bicycles and tricycles. In total there are 7 cycles (cycles includes both bicycles and tricycles) and 19 wheels. Determine how many of each there are, if a bicycle has two wheels and a tricycle has three wheels.

##### Solution
1. Step 1. Identify what is required :

The number of bicycles and the number of tricycles are required.

2. Step 2. Set up the necessary equations :

If bb is the number of bicycles and tt is the number of tricycles, then:

b + t = 7 2 b + 3 t = 19 b + t = 7 2 b + 3 t = 19
(80)
3. Step 3. Solve the system of simultaneous equations using substitution. :
b = 7 - t Into second equation: 2 ( 7 - t ) + 3 t = 19 14 - 2 t + 3 t = 19 t = 5 Into first equation: : b = 7 - 5 = 2 b = 7 - t Into second equation: 2 ( 7 - t ) + 3 t = 19 14 - 2 t + 3 t = 19 t = 5 Into first equation: : b = 7 - 5 = 2
(81)
4. Step 4. Check solution by substituting into original system of equations. :
2 + 5 = 7 2 ( 2 ) + 3 ( 5 ) = 4 + 15 = 19 2 + 5 = 7 2 ( 2 ) + 3 ( 5 ) = 4 + 15 = 19
(82)

#### Simultaneous Equations


2. Solve algebraically: 15c+11d-132=015c+11d-132=0, 2c+3d-59=02c+3d-59=0

3. Solve algebraically: -18e-18+3f=0-18e-18+3f=0, e-4f+47=0e-4f+47=0

4. Solve graphically: x+2y=7x+2y=7, x+y=0x+y=0


## Literal equations

A literal equation is one that has several letters or variables. Examples include the area of a circle (A=πr2A=πr2) and the formula for speed (s=dts=dt). In this section you will learn how to solve literal equations in terms of one variable. To do this, you will use the principles you have learnt about solving equations and apply them to rearranging literal equations. Solving literal equations is also known as changing the subject of the formula.

You should bear the following in mind when solving literal equations:

• We isolate the unknown by asking what is joined to it, how this is joined and we do the opposite operation (to both sides as a whole).
• If the unknown variable is in two or more terms, then we take it out as a common factor.
• If we have to take the square root of both sides, remember that there will be a positive and a negative answer.
• If the unknown variable is in the denominator, then we find the lowest common denominator (LCD), multiply both sides by LCD and then continue to solve the problem.

### Exercise 17: Solving literal equations - 1

The area of a triangle is A=12b×hA=12b×h. What is the height of the triangle in terms of the base and area?

#### Solution

1. Step 1. Rearrange the equation: We rearrange the equation so that the height is on one side of the equals sign and the rest of the variables are on the other side of the equation.
A = 12 b×h 2A = b×h ( multiply both sides by 2 ) 2Ab = h ( divide both sides by b ) A = 12 b×h 2A = b×h ( multiply both sides by 2 ) 2Ab = h ( divide both sides by b )
(83)
2. Step 2. Write the final answer:

The height of a triangle is given by: h=2Abh=2Ab

### Solving literal equations

1. Solve for t: v=u+at v=u+at
2. Solve for x: ax-bx=c ax-bx=c
3. Solve for x: 1b+2bx=21b+2bx=2

## Mathematical Models (Not in CAPS - included for completeness)

### Introduction

Tom and Jane are friends. Tom picked up Jane's Physics test paper, but will not tell Jane what her marks are. He knows that Jane hates maths so he decided to tease her. Tom says: “I have 2 marks more than you do and the sum of both our marks is equal to 14. How much did we get?”

Let's help Jane find out what her marks are. We have two unknowns, Tom's mark (which we shall call tt) and Jane's mark (which we shall call jj). Tom has 2 more marks than Jane. Therefore,

t = j + 2 t = j + 2
(84)

Also, both marks add up to 14. Therefore,

t + j = 14 t + j = 14
(85)

The two equations make up a set of linear (because the highest power is one) simultaneous equations, which we know how to solve! Substitute for tt in the second equation to get:

t + j = 14 j + 2 + j = 14 2 j + 2 = 14 2 ( j + 1 ) = 14 j + 1 = 7 j = 7 - 1 = 6 t + j = 14 j + 2 + j = 14 2 j + 2 = 14 2 ( j + 1 ) = 14 j + 1 = 7 j = 7 - 1 = 6
(86)

Then,

t = j + 2 = 6 + 2 = 8 t = j + 2 = 6 + 2 = 8
(87)

So, we see that Tom scored 8 on his test and Jane scored 6.

This problem is an example of a simple mathematical model. We took a problem and we were able to write a set of equations that represented the problem mathematically. The solution of the equations then gave the solution to the problem.

### Problem Solving Strategy

The purpose of this section is to teach you the skills that you need to be able to take a problem and formulate it mathematically in order to solve it. The general steps to follow are:

1. Read ALL of the question !
2. Find out what is requested.
3. Use a variable(s) to denote the unknown quantity/quantities that has/have been requested e.g., xx.
4. Rewrite the information given in terms of the variable(s). That is, translate the words into algebraic expressions.
5. Set up an equation or set of equations (i.e. a mathematical sentence or model) to solve the required variable.
6. Solve the equation algebraically to find the result.

### Application of Mathematical Modelling

#### Exercise 18: Mathematical Modelling: Two variables

Three rulers and two pens have a total cost of R 21,00. One ruler and one pen have a total cost of R 8,00. How much does a ruler costs on its own and how much does a pen cost on its own?

##### Solution
1. Step 1. Translate the problem using variables :

Let the cost of one ruler be xx rand and the cost of one pen be yy rand.

2. Step 2. Rewrite the information in terms of the variables :
3 x + 2 y = 21 x + y = 8 3 x + 2 y = 21 x + y = 8
(88)
3. Step 3. Solve the equations simultaneously :

First solve the second equation for yy:

y = 8 - x y = 8 - x
(89)

and substitute the result into the first equation:

3 x + 2 ( 8 - x ) = 21 3 x + 16 - 2 x = 21 x = 5 3 x + 2 ( 8 - x ) = 21 3 x + 16 - 2 x = 21 x = 5
(90)

therefore

y = 8 - 5 y = 3 y = 8 - 5 y = 3
(91)
4. Step 4. Present the final answers :

One ruler costs R 5,00 and one pen costs R 3,00.

#### Exercise 19: Mathematical Modelling: One variable

A fruit shake costs R2,00 more than a chocolate milkshake. If three fruit shakes and 5 chocolate milkshakes cost R78,00, determine the individual prices.

##### Solution
1. Step 1. Summarise the information in a table :

Let the price of a chocolate milkshake be xx and the price of a fruitshake be yy.

 Price number Total Fruit y y 3 3 y 3 y Chocolate x x 5 5 x 5 x
2. Step 2. Set up a pair of algebraic equations :
3 y + 5 x = 78 3 y + 5 x = 78
(92)

y=x+2y=x+2

3. Step 3. Solve the equations :
3 ( x + 2 ) + 5 x = 78 3 x + 6 + 5 x = 78 8 x = 72 x = 9 y = x+2 = 9 + 2 = 11 3 ( x + 2 ) + 5 x = 78 3 x + 6 + 5 x = 78 8 x = 72 x = 9 y = x+2 = 9 + 2 = 11
(93)
4. Step 4. Present the final answer :

One chocolate milkshake costs R 9,00 and one Fruitshake costs R 11,00

#### Mathematical Models

1. Stephen has 1 l of a mixture containing 69% of salt. How much water must Stephen add to make the mixture 50% salt? Write your answer as a fraction of a litre.

2. The diagonal of a rectangle is 25 cm more than its width. The length of the rectangle is 17 cm more than its width. What are the dimensions of the rectangle?

3. The sum of 27 and 12 is 73 more than an unknown number. Find the unknown number.

4. The two smaller angles in a right-angled triangle are in the ratio of 1:2. What are the sizes of the two angles?

5. George owns a bakery that specialises in wedding cakes. For each wedding cake, it costs George R150 for ingredients, R50 for overhead, and R5 for advertising. George's wedding cakes cost R400 each. As a percentage of George's costs, how much profit does he make for each cake sold?

6. If 4 times a number is increased by 7, the result is 15 less than the square of the number. Find the numbers that satisfy this statement, by formulating an equation and then solving it.

7. The length of a rectangle is 2 cm more than the width of the rectangle. The perimeter of the rectangle is 20 cm. Find the length and the width of the rectangle.


### Summary

• A linear equation is an equation where the power of the variable(letter, e.g. xx) is 1(one). A linear equation has at most one solution
• A quadratic equation is an equation where the power of the variable is at most 2. A quadratic equation has at most two solutions
• Exponential equations generally have the unknown variable as the power. The general form of an exponential equation is: ka(x+p)=mka(x+p)=m
• A linear inequality is similar to a linear equation and has the power of the variable equal to 1. When you divide or multiply both sides of an inequality by any number with a minus sign, the direction of the inequality changes. You can solve linear inequalities using the same methods used for linear equations
• When two unknown variables need to be solved for, two equations are required and these equations are known as simultaneous equations. There are two ways to solve linear simultaneous equations: graphical solutions and algebraic solutions. To solve graphically you draw the graph of each equation and the solution will be the co-ordinates of the point of intersection. To solve algebraically you solve equation one, for variable one and then substitute that solution into equation two and solve for variable two.
• Literal equations are equations where you have several letters (variables) and you rearrange the equation to find the solution in terms of one letter (variable)
• Mathematical modelling is where we take a problem and we write a set of equations that represent the problem mathematically. The solution of the equations then gives the solution to the problem.

### End of Chapter Exercises

1. What are the roots of the quadratic equation x2-3x+2=0x2-3x+2=0

?

2. What are the solutions to the equation x2+x=6x2+x=6

?

3. In the equation y=2x2-5x-18y=2x2-5x-18, which is a value of xx when y=0y=0

?

4. Manuel has 5 more CDs than Pedro has. Bob has twice as many CDs as Manuel has. Altogether the boys have 63 CDs. Find how many CDs each person has.

5. Seven-eighths of a certain number is 5 more than one-third of the number. Find the number.

6. A man runs to a telephone and back in 15 minutes. His speed on the way to the telephone is 5 m/s and his speed on the way back is 4 m/s. Find the distance to the telephone.

7. Solve the inequality and then answer the questions: x3-14>14-x4x3-14>14-x4
1. If xRxR, write the solution in interval notation.
2. if xZxZ and x<51x<51, write the solution as a set of integers.

8. Solve for aa: 1-a2-2-a3>11-a2-2-a3>1

9. Solve for xx: x-1=42xx-1=42x

10. Solve for xx and yy: 7x+3y=137x+3y=13 and 2x-3y=-42x-3y=-4


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