**Simultaneous equations **

A pair of linked equations with unknown numbers represented by the same two letters in each equation.

Example

2x + 3y = 11

5x + 2y = 18

To solve a pair of simultaneous equations, find the values for the unknowns that fit **both** equations.

Simultaneous equations can be solved graphically or algebraically.

**Solving graphically**

Plot the graphs of both equations on the same set of axes.

The point(s) where the graphs cross are the points where the x and y values satisfy both equations.

The co-ordinates of the point(s) where the graphs cross are your solutions.

**Solving algebraically**

There are two methods for solving simultaneous equations using algebra, **elimination** and **substitution**. Both methods involve reducing the problem to an equation with only one unknown, which can be solved. Once one unknown has been found, the other can also be found.

**Elimination**

**Example 1**

3x – y = 1 The first equation has a – y term

x + y = 3 and the second a + y term

3x + x – y + y = 1 + 3 Adding the two equations together will eliminate the y terms

4x = 4 Simplify by combining like terms, the result is an equation with only one unknown

x = 1 Solve to find x

1 + y = 3 Put your value for x into one of the

original equations

y = 2 Solve to find y

The solutions are x = 1 and y = 2

**Example 2**

2x + 3y = 9 (1) Here the coefficients (numbers in

x + 4y = 7 (2) front of) x and y are both different

Label your equations (1) and (2) (or A and B if you prefer).

We want a pair of equations where the coefficient of one of the unknowns is the same in both equations.

2x + 8y = 14 (3) Equation (2) multiplied by 2, label (3)

2x – 2x + 8y – 3y = 14 – 9 Equation (3) minus equation (1)

5y = 5 Simplify by combining like terms, the result is an equation with only one unknown

y = 1 Solve to find y

2x + 3 = 9 Put your value for y into one of the original equations

2x = 6

x = 3 Solve to find x

The solutions are x = 3 and y = 1

**Elimination method summary – six steps to follow **

§ Rearrange both equations into the form ax + by = c where a, b and c are numbers, if the equations are not already in this form. Label the equations (1) and (2).

§ Match up the coefficients (numbers in front of) either the x’s or the y’s. To do this you may need to multiply either one or both of the equations by suitable numbers. Label the new equation(s) (3) (and (4)).

§ Add or subtract the equations in order to eliminate the unknowns with the same coefficient. If the coefficients have the same sign (are both positive or both negative) subtract – as in Example 2 above. If the coefficients have opposite signs (one positive and one negative) add – as in Example 1 above.

§ Solve the resulting equation to find the value of the remaining unknown (letter).

§ Substitute the value obtained into one of the original equations, and solve to find the value of the other unknown (letter).

§ As a check, substitute both values obtained into the original equation you didn’t use in the previous step. Make sure the values work correctly – if not, then you know something has gone wrong. Check through your work to find the mistake.

**Substitution**

**Example 1**

5x + 6y = 34

y = x + 2 This second equation gives y in terms of x

5x + 6(x + 2) = 34 Substitute y in terms of x into the first equation

5x + 6x + 12 = 34 Multiply out brackets

11x + 12 = 34 Round up terms

11x = 22

x = 11 Solve to find x

55 + 6y = 34 Substitute the value of x back in

6y = -21

y = -3.5 Solve to find y

The solutions are x = 11 and y = -3.5

**Example 2**

2x^{2} – y = 5

x + y = 1 Rearrange this second equation to

y = 1 – x give y in terms of x

2x^{2} – (1 – x) = 5 Substitute y in terms of x into the first equation

2x^{2} – 1 + x = 5 Remove brackets, taking care with signs

2x^{2} + x – 6 = 0 Round up terms and rearrange

(2x – 3)(x + 2) = 0 Factorise the quadratic

x = 1.5 or x = -2 Solve for x

1.5 + y = 1 Substitute one value of x back in to one of the original equations

y = -0.5 Solve to find the corresponding y

value

-2 + y = 1 Substitute the other x value back in

y = 3 Solve to find the corresponding y

The solutions are x = 1.5, y = -0.5 and x = -2, y = 3

**Substitution method summary – seven steps to follow**

(This summary applies to pairs of simultaneous equations where one has a quadratic expression, as in Example 2 above. Example 1 type equations should be straightforward)

§ Rearrange one of the equations so that you have one unknown in terms of the other, like in Example 2 above where x + y = 1 becomes y = 1 – x

§ Substitute this into the quadratic (the equation with a squared term), in Example 2 you get 2x^{2} – (1 – x) = 5 (The (1-x) has replaced the y)

§ Simplify and rearrange the result to get a quadratic equation with zero on one side, such as 2x^{2} + x – 6 = 0

§ Solve the quadratic equation by factorising, completing the square or using the quadratic formula to find the value(s) of one unknown

§ Substitute the first solution back in to one of the original equations to find the corresponding value of the second unknown

§ Substitute the second solution back into the same original equation used in the previous step to find the corresponding value of the second unknown

§ Substitute both pairs of solutions back into the other original equation (one pair at a time!) to check that they work

§ Write the solutions out clearly in the correct pairing, for example

The solutions are x = 1.5, y = -0.5 and x = -2, y = 3

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