GCSE Maths

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Algebra
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Linear Simultaneous Equations
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Linear Simultaneous Equations

GCSE Maths Algebra Linear Simultaneous Equations

Simultaneous equations are a set of equations with multiple variables that are solved together, such that the solution satisfies all the equations in the system. Linear simultaneous equations are equations where all the variables are raised to the power of 1 and are not multiplied by other variables.

There are two types of linear simultaneous equations you will typically come across:

  • Linear simultaneous equations with two variables (x, y)
  • Linear simultaneous equations with three variables (x, y, z)

Methods for Solving Linear Simultaneous Equations:

The two main methods we’ll look at for solving linear simultaneous equations are the substitution method and the elimination method.

Substitution method

The substitution method is a technique used to solve simultaneous equations by expressing one variable in terms of the other variable, and then substituting this expression into the other equation.

This process reduces the number of variables in the equation, allowing us to solve for the remaining variable. Once we have found the value of one variable, we can then substitute it back into the original expression to find the value of the other variable.

For example, consider the following system of linear simultaneous equations:

x + y = 5

2x – y = 1

Using the substitution method, follow these steps:

Step 1: Solve one equation for one variable in terms of the other variable. We will solve the first equation for x in terms of y:

  • x = 5 – y

Step 2: Substitute the expression from step 1 into the other equation. Now, substitute x = 5 – y into the second equation:

  • 2(5 – y) – y = 1

Step 3: Solve the resulting equation for the remaining variable. We solve the equation for y:

  • 10 – 2y – y = 1
  • 3y = 9
  • y = 3

Step 4: Substitute the value from step 3 back into the expression from step 1 to find the other variable. We substitute y = 3 back into the expression for x:

x = 5 – 3

x = 2

The solution to the system of simultaneous equations is (x, y) = (2, 3).

Elimination method

The elimination method is a technique used to solve simultaneous equations by manipulating the equations in such a way that one of the variables can be eliminated by adding or subtracting the equations.

This is typically done by multiplying one or both equations by necessary factors to make the coefficients of one of the variables the same or additive inverses.

  • Additive inverses are pairs of numbers that, when added together, result in a sum of zero.

Once one variable is eliminated, the remaining equation can be solved for the other variable. Finally, the value of the remaining variable can be substituted back into one of the original equations to find the value of the eliminated variable.

For example, consider the following simultaneous equations:

4x + 3y = 1

2x – 3y = 11

Using the elimination method, follow these steps:

Step 1: Multiply one or both equations by a necessary factor to make the coefficients of one of the variables the same or additive inverses. In this case, we don’t need to multiply any equation because the coefficients of y are already additive inverses (3 and –3).

Step 2: Add or subtract the equations to eliminate one variable. We add both equations to eliminate the y variable:

  4x + 3y = 1
+ 2x – 3y = 11
----------------
  6x      = 12

Step 3: Solve the resulting equation for the remaining variable. We solve the equation for x:

  • 6x = 12
  • x = 2

Step 4: Substitute the value from step 3 back into one of the original equations to find the other variable. We substitute x = 2 back into the first equation:

  • 4(2) + 3y = 1
  • 8 + 3y = 1
  • 3y = –7
  • y = \frac{-7}{3}

The solution to the simultaneous equations is (x, y) = (2, \frac{-7}{3}).

Examples

Example 1:

Solve the following linear simultaneous equations with three variables using either the substitution or elimination method:

3x + 4y = 25

5x – 2y = 7

We will use the elimination method. First, multiply both equations to make the coefficients of y additive inverses:

Multiply the first equation by 2:

• 6x + 8y = 50

Multiply the second equation by 4:

• 20x – 8y = 28

Now, add the two equations:

     6x + 8y = 50
+ 20x – 8y = 28
– – – – – – – – – – – – – 
     26x         = 78

Divide by 26 to solve for x:

x = \frac{78}{26}

x = 3

Substitute the value of x back into one of the original equations, for example, equation 1:

3(3) + 4y = 25

9 + 4y = 25

Subtract 9 from both sides:

4y = 16

Divide by 4 to solve for y:

y = \frac{16}{4}

y = 4

Solution (x, y) = (3, 4)

Example 2:

Solve the following simultaneous equations using the substitution method:

3x – 2y = 7

x + y = 3

1. Solve for x in the second equation: x = 3 – y

2. Substitute into the first equation: 3(3 – y) – 2y = 7

3. Solve for y:

9 – 3y – 2y = 7

• 5y = 2

• y = \frac{2}{5}

4. Substitute y back into the expression for x:

x = 3 - \frac{2}{5}

x = \frac{15}{5} - \frac{2}{5}

x = \frac{13}{5}

Solution (x, y) = (\frac{13}{5}, \frac{2}{5})

Example 3:

Solve the following linear simultaneous equations using the elimination method:

4x + 3y = 1

2x – 3y = 11

1. The coefficients of y are already additive inverses (3 and -3), so you can just add the two equations as they are:

    4x + 3y = 1
+  2x – 3y = 11
– – – – – – – – – – – – – 
    6x         = 12

2. Solve for x:

x = \frac{12}{6}

x = 2

3. Substitute x back into one of the original equations, for example, the first one:

4(2) + 3y = 1

4. Solve for y:

8 + 3y = 1

3y = –7

y = \frac{-7}{3}

So the solution is (x, y) = (2, \frac{-7}{3}).

Example 4:

Solve the following linear simultaneous equations with three variables using either the substitution or elimination method:

x + y + z = 6

2x – y + z = 4

x + 2y – z = 2

1. Solve the first equation for x:

x = 6 – y – z

2. Substitute into the second and third equations:

Second:

2(6 – y – z) – y + z = 4

12 – 2y – 2z – y + z = 4

–3y – z = –8

Third:

(6 – y – z) + 2y – z = 2

6 – y – z + 2y – z = 2

y – 2z = –4

Now we have two equations with two variables:

–3y – z = –8

y – 2z = –4

3. Multiply the second equation by 3 and add the two equations:

3(y – 2z) = 3(-4)

3y – 6z = –12

Add the two equations:

    –3y –   z = –8
+   3y – 6z = –12
– – – – – – – – – – – – – 
           -7z = -20

4. Substitute into the first equation: –3(2z – 4) – z = –8

–6z + 12 – z = –8

–7z = –20

z = \frac{20}{7}

5. Substitute z back into the expression for y:

y = 2(\frac{20}{7}) - 4

y = \frac{40}{7} - 4

y = \frac{4}{7}

6. Substitute y and z back into the expression for x:

x = 6 - \frac{4}{7} - \frac{20}{7}

x = \frac{42}{7} - \frac{4}{7} - \frac{20}{7}

x = \frac{18}{7}

Solution (x, y, z) = (\frac{18}{7}, \frac{4}{7}, \frac{20}{7})

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