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Equations and inequalities: 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
(1)

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 1
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
(2)

Draw the graphs of the two equations in Equation 2.

Figure 2
Figure 2 (MG10C10_006.png)

The intersection of the two graphs is (2,1)(2,1). So the solution to the system of simultaneous equations in Equation 2 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
(3)

Exercise 1: 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
(4)
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
    (5)

    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
    (6)

    Figure 3
    Figure 3 (MG10C10_007.png)

  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
    (7)

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 1, we first solve the first equation for xx:

x = 1 2 - y x = 1 2 - y
(8)

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
(9)
∴ x = 1 2 - y = 1 2 - ( - 1 10 ) = 6 10 = 3 5 ∴ x = 1 2 - y = 1 2 - ( - 1 10 ) = 6 10 = 3 5
(10)

The solution for the system of simultaneous equations Equation 1 is:

x = 3 5 y = - 1 10 x = 3 5 y = - 1 10
(11)

Exercise 2: 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
(12)
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
    (13)
  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
    (14)
  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
    (15)
  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
    (16)

Exercise 3: 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
    (17)
  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
    (18)
  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
    (19)

Simultaneous Equations

  1. Solve graphically and confirm your answer algebraically: 3a-2b7=03a-2b7=0 , a-4b+1=0a-4b+1=0
     
    Click here for the solution
  2. Solve algebraically: 15c+11d-132=015c+11d-132=0, 2c+3d-59=02c+3d-59=0
     
    Click here for the solution
  3. Solve algebraically: -18e-18+3f=0-18e-18+3f=0, e-4f+47=0e-4f+47=0
     
    Click here for the solution
  4. Solve graphically: x+2y=7x+2y=7, x+y=0x+y=0
     
    Click here for the solution

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