Connexions

You are here: Home » Content » Siyavula textbooks: Grade 10 Maths [NCS] » Solving linear equations

Lenses

What is a lens?

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags?

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

Affiliated with (What does "Affiliated with" mean?)

This content is either by members of the organizations listed or about topics related to the organizations listed. Click each link to see a list of all content affiliated with the organization.
• Siyavula: Mathematics

This collection is included inLens: Siyavula Textbooks: Maths
By: Free High School Science Texts Project

Click the "Siyavula: Mathematics" link to see all content affiliated with them.

Click the tag icon to display tags associated with this content.

• Bookshare

This collection is included inLens: Bookshare's Lens
By: Bookshare - A Benetech Initiative

"Accessible versions of this collection are available at Bookshare. DAISY and BRF provided."

Click the "Bookshare" link to see all content affiliated with them.

• FETMaths

This module and collection are included inLens: Siyavula: Mathematics (Gr. 10-12)
By: Siyavula

Module Review Status: In Review
Collection Review Status: In Review

Click the "FETMaths" link to see all content affiliated with them.

Click the tag icon to display tags associated with this content.

Recently Viewed

This feature requires Javascript to be enabled.

Tags

(What is a tag?)

These tags come from the endorsement, affiliation, and other lenses that include this content.

Inside Collection (Textbook):

Solving linear equations

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 review of past work (section on 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 : Strategy for Solving Equations

In the following, identify what is wrong.

4 x - 8 = 3 ( x - 2 ) 4 ( x - 2 ) = 3 ( x - 2 ) 4 ( x - 2 ) ( x - 2 ) = 3 ( x - 2 ) ( x - 2 ) 4 = 3 4 x - 8 = 3 ( x - 2 ) 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 a possible ) 2 x + 2 = 1 2 x = 1 - 2 ( like terms together ) 2 x = - 1 ( simplified as much a 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:

L H S = 2 x + 2 = 2 ( - 1 2 ) + 2 = - 1 + 2 = 1 a n d R H S = 1 L H S = 2 x + 2 = 2 ( - 1 2 ) + 2 = - 1 + 2 = 1 a n d R H S = 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 Equations

The general steps to solve equations are:

1. Expand (Remove) all brackets.
2. "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 all like terms together and simplify as much as possible.
4. Factorise if necessary.
5. Find the solution.
6. 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 grouping like terms and then simplifying.

3. Step 3. Solve the problem :
4 - x = 4 - x = 4 - 4 ( move all constant terms ( numbers ) to the RHS ( right hand side ) ) - x = 0 ( group like terms together ) - x = 0 ( simplify grouped terms ) - x = 0 x = 0 4 - x = 4 - x = 4 - 4 ( move all constant terms ( numbers ) to the RHS ( right hand side ) ) - x = 0 ( group like terms together ) - x = 0 ( simplify grouped terms ) - x = 0 x = 0
(8)
4. Step 4. Check the answer :

Substitute solution into original equation:

4 - 0 = 4 4 - 0 = 4
(9)
4 = 4 4 = 4
(10)

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 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 move all terms with x to the LHS and all constant terms to the RHS of the = ( 8 x - 4 x + 6 x ) = ( 4 + 36 ) ( group like terms together ) 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 move all terms with x to the LHS and all constant terms to the RHS of the = ( 8 x - 4 x + 6 x ) = ( 4 + 36 ) ( group like terms together ) 10 x = 40 ( simplify grouped terms ) 10 10 x = 40 10 ( divide both sides by 10 ) x = 4
(11)
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
(12)

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 ( remove / expand brackets ) - x - 6 x = 2 - 2 move all terms containing x to the LHS and all constant terms ( numbers ) to the RHS . - 7 x = 0 ( simplify grouped terms ) x = 0 ÷ ( - 7 ) t h e r e f o r e 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 ( remove / expand brackets ) - x - 6 x = 2 - 2 move all terms containing x to the LHS and all constant terms ( numbers ) to the RHS . - 7 x = 0 ( simplify grouped terms ) x = 0 ÷ ( - 7 ) t h e r e f o r e x = 0 zero divided by any number is 0
(13)
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
(14)

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 ( move all terms with x to the LHS and all constant terms to the RHS of the = ) - 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 ( move all terms with x to the LHS and all constant terms to the RHS of the = ) - 17 x = 24 ( simplify grouped terms ) - 17 - 17 x = 24 - 17 ( divide both sides by - 17 ) x = - 24 17
(15)
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
(16)

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 ww: -3w=0-3w=0

3. Solve for zz: 4z=164z=16

4. Solve for tt: 12t+0=14412t+0=144

5. Solve for xx: 7+5x=627+5x=62

6. Solve for yy: 55=5y+3455=5y+34

7. Solve for zz: 5z=3z+455z=3z+45

8. Solve for aa: 23a-12=6+2a23a-12=6+2a

9. Solve for bb: 12-6b+34b=2b-24-6412-6b+34b=2b-24-64

10. Solve for cc: 6c+3c=4-5(2c-3)6c+3c=4-5(2c-3)

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

12. Solve for qq: 4q=16244q=1624

13. Solve for qq: 41=q241=q2

14. Solve for rr: -(-16-r)=13r-1-(-16-r)=13r-1

15. Solve for dd: 6d-2+2d=-2+4d+86d-2+2d=-2+4d+8

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

17. Solve for vv: 3v+16=4v-103v+16=4v-10

18. Solve for kk: 10k+5+0=-2k+-3k+8010k+5+0=-2k+-3k+80

19. Solve for jj: 8(j-4)=5(j-4)8(j-4)=5(j-4)

20. Solve for mm: 6=6(m+7)+5m6=6(m+7)+5m


Content actions

PDF | EPUB (?)

What is an EPUB file?

EPUB is an electronic book format that can be read on a variety of mobile devices.

PDF | EPUB (?)

What is an EPUB file?

EPUB is an electronic book format that can be read on a variety of mobile devices.

Collection to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags?

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks

Module to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags?

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks