Simultaneous equations Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Simultaneous equations questions. See exactly how to solve problems on simultaneous equations, elimination, linear systems, graphical interpretation.

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A-Level70 questionsStep-by-step solutions

Question 1

3 markseasy

Solve the simultaneous equations x + y = 5 x - y = 1

Worked solution

Label the two equations
$x + y = 5 \quad (1)\\ x - y = 1 \quad (2)$
We label them (1) and (2) so we can refer to each equation clearly. The plan is to eliminate one letter so we are left with a single equation in one unknown.
Add the equations
$(1')+(2'):\quad 2 x = 6$
Because the $y$ -terms are equal and opposite we ADD, which cancels $y$ . This leaves one equation with only $x$ in it.
Solve for $x$
$x = \dfrac{6}{2} = 3$
Dividing both sides by the number in front of $x$ isolates $x$ . This is the same rearranging skill used all through algebra.
Substitute back to find $y$
$y + 3 = 5 \Rightarrow y = 2 \Rightarrow y = 2$
Put the value of $x$ into equation (1) and solve the small equation that remains. Using the original equation keeps the numbers exact.
State the solution
$x = 3,\ y = 2$
This ordered pair is where the two straight lines cross on a graph, so it is the single point that satisfies both equations at once.

Answer

x = 3,\ y = 2

Question 2

3 markseasy

Solve the simultaneous equations 2 x + y = 7 x - y = 2

Worked solution

Label the two equations
$2 x + y = 7 \quad (1)\\ x - y = 2 \quad (2)$
We label them (1) and (2) so we can refer to each equation clearly. The plan is to eliminate one letter so we are left with a single equation in one unknown.
Add the equations
$(1')+(2'):\quad 3 x = 9$
Because the $y$ -terms are equal and opposite we ADD, which cancels $y$ . This leaves one equation with only $x$ in it.
Solve for $x$
$x = \dfrac{9}{3} = 3$
Dividing both sides by the number in front of $x$ isolates $x$ . This is the same rearranging skill used all through algebra.
Substitute back to find $y$
$y + 6 = 7 \Rightarrow y = 1 \Rightarrow y = 1$
Put the value of $x$ into equation (1) and solve the small equation that remains. Using the original equation keeps the numbers exact.
State the solution
$x = 3,\ y = 1$
This ordered pair is where the two straight lines cross on a graph, so it is the single point that satisfies both equations at once.

Answer

x = 3,\ y = 1

Question 3

3 markseasy

Solve the simultaneous equations 3 x + 2 y = 12 x + 2 y = 8

Worked solution

Label the two equations
$3 x + 2 y = 12 \quad (1)\\ x + 2 y = 8 \quad (2)$
We label them (1) and (2) so we can refer to each equation clearly. The plan is to eliminate one letter so we are left with a single equation in one unknown.
Subtract the equations
$(1')-(2'):\quad 2 x = 4$
Because the $y$ -terms are now identical we SUBTRACT, which cancels $y$ . This leaves one equation with only $x$ in it.
Solve for $x$
$x = \dfrac{4}{2} = 2$
Dividing both sides by the number in front of $x$ isolates $x$ . This is the same rearranging skill used all through algebra.
Substitute back to find $y$
$2 y + 6 = 12 \Rightarrow 2 y = 6 \Rightarrow y = 3$
Put the value of $x$ into equation (1) and solve the small equation that remains. Using the original equation keeps the numbers exact.
State the solution
$x = 2,\ y = 3$
This ordered pair is where the two straight lines cross on a graph, so it is the single point that satisfies both equations at once.

Answer

x = 2,\ y = 3

Question 4

3 markseasy

Solve the simultaneous equations x + 2 y = 11 x - y = 2

Worked solution

Label the two equations
$x + 2 y = 11 \quad (1)\\ x - y = 2 \quad (2)$
We label them (1) and (2) so we can refer to each equation clearly. The plan is to eliminate one letter so we are left with a single equation in one unknown.
Match the $y$ -coefficients
$(1)\times 1:\ x + 2 y = 11\\ (2)\times 2:\ 2 x - 2 y = 4$
To eliminate $y$ the $y$ -terms must match. We multiply (1) by 1 and (2) by 2 so both have $y$ -coefficient of the same size. Multiplying every term keeps each equation balanced.
Add the equations
$(1')+(2'):\quad 3 x = 15$
Because the $y$ -terms are equal and opposite we ADD, which cancels $y$ . This leaves one equation with only $x$ in it.
Solve for $x$
$x = \dfrac{15}{3} = 5$
Dividing both sides by the number in front of $x$ isolates $x$ . This is the same rearranging skill used all through algebra.
Substitute back to find $y$
$2 y + 5 = 11 \Rightarrow 2 y = 6 \Rightarrow y = 3$
Put the value of $x$ into equation (1) and solve the small equation that remains. Using the original equation keeps the numbers exact.
State the solution
$x = 5,\ y = 3$
This ordered pair is where the two straight lines cross on a graph, so it is the single point that satisfies both equations at once.

Answer

x = 5,\ y = 3

Question 5

3 markseasy

Solve the simultaneous equations 2 x + 3 y = 12 x - y = 1

Worked solution

Label the two equations
$2 x + 3 y = 12 \quad (1)\\ x - y = 1 \quad (2)$
We label them (1) and (2) so we can refer to each equation clearly. The plan is to eliminate one letter so we are left with a single equation in one unknown.
Match the $y$ -coefficients
$(1)\times 1:\ 2 x + 3 y = 12\\ (2)\times 3:\ 3 x - 3 y = 3$
To eliminate $y$ the $y$ -terms must match. We multiply (1) by 1 and (2) by 3 so both have $y$ -coefficient of the same size. Multiplying every term keeps each equation balanced.
Add the equations
$(1')+(2'):\quad 5 x = 15$
Because the $y$ -terms are equal and opposite we ADD, which cancels $y$ . This leaves one equation with only $x$ in it.
Solve for $x$
$x = \dfrac{15}{5} = 3$
Dividing both sides by the number in front of $x$ isolates $x$ . This is the same rearranging skill used all through algebra.
Substitute back to find $y$
$3 y + 6 = 12 \Rightarrow 3 y = 6 \Rightarrow y = 2$
Put the value of $x$ into equation (1) and solve the small equation that remains. Using the original equation keeps the numbers exact.
State the solution
$x = 3,\ y = 2$
This ordered pair is where the two straight lines cross on a graph, so it is the single point that satisfies both equations at once.

Answer

x = 3,\ y = 2

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