Hard GCSE Simultaneous equations (linear) Questions

Challenging, exam-style GCSE Simultaneous equations (linear) questions with worked solutions. Stretch yourself on the hardest elimination, scaling both equations, negative solution, forming equations problems.

eliminationscaling both equationsnegative solutionforming equationsmoney problemgraphical solution
GCSE Foundation34 questionsStep-by-step solutions
Question 1
5 markschallenging
Marta says the solution of the simultaneous equations 3x+2y=133x + 2y = 13 and x−y=1x - y = 1 is x=2x = 2, y=3y = 3. Explain why Marta is wrong, and give the correct solution.
Show worked solution

Worked solution

  1. Test Marta's values in equation (1)

    3(2)+2(3)=6+6=12≠133(2) + 2(3) = 6 + 6 = 12 \ne 13

    Substituting x=2x = 2 and y=3y = 3 into 3x + 2y gives 12, not 13, so her pair fails the first equation.

  2. Test Marta's values in equation (2)

    (2)−(3)=−1≠1(2) - (3) = -1 \ne 1

    It fails the second equation as well: x - y should be 1, but she gets -1. The sign being reversed is a clue that the values have been swapped.

  3. Label the two equations

    (1)  3x+2y=13(2)  x−y=1(1)\ \ 3x + 2y = 13 \qquad (2)\ \ x - y = 1

    Numbering the equations makes it easy to say exactly what you are doing to each one.

  4. Decide which variable to eliminate

    y-coefficients: 2 and −1⇒match them at 2\text{y-coefficients: } 2 \text{ and } -1 \Rightarrow \text{match them at } 2

    To eliminate y the two y-coefficients must be the same size. The lowest common multiple of 2 and 1 is 2, so scale each equation up to that.

  5. Keep equation (1) as it is

    (3)  3x+2y=13(3)\ \ 3x + 2y = 13

    The y-coefficient here is already the right size, so equation (1) does not need scaling.

  6. Multiply equation (2) by 2

    (4)  2x−2y=2(4)\ \ 2x - 2y = 2

    Multiply every term — both sides — by 2. The equation still has the same solutions, but the y-coefficient is now 2 in size.

  7. Add the scaled equations

    (3)+(4):(3x+2y)+(2x−2y)=13+2(3) + (4): \quad (3x + 2y) + (2x - 2y) = 13 + 2

    The y-terms are now the same size and the signs are opposite, so adding makes them cancel. Adding the two equations removes y completely.

  8. Eliminate y

    (3)+(4):5x=15(3) + (4): \quad 5x = 15

    The y-terms cancel, leaving one equation with only x in it.

  9. Solve for x

    x=155=3x = \frac{15}{5} = 3

    Divide both sides by 5.

  10. Substitute x=3x = 3 into equation (1)

    3(3)+2y=133(3) + 2y = 13

    Put the value you have just found back into one of the original equations to find y.

  11. Simplify

    9+2y=139 + 2y = 13

    Work out the number part first, then the equation has only one unknown left.

  12. Solve for y

    2y=13−9=4⇒y=22y = 13 - 9 = 4 \Rightarrow y = 2

    Move the number to the other side, then divide by 2.

  13. Check in equation (2)

    (3)−(2)=3−2=1 ✓(3) - (2) = 3 - 2 = 1 \ \checkmark

    Substituting both values into the equation you did not use for the back-substitution confirms the solution is right. Always do this check.

  14. State the solution

    x=3, y=2x = 3,\ y = 2

    These values satisfy both equations at once — they are the coordinates of the point where the two lines cross.

  15. Explain the error

    Correct: x=3, y=2(Marta had them the wrong way round)\text{Correct: } x = 3,\ y = 2 \quad \text{(Marta had them the wrong way round)}

    Marta found the right pair of numbers but wrote them against the wrong letters. Always state which value belongs to x and which to y, then check both equations.

Answer
x=3, y=2x = 3,\ y = 2
Question 2
5 markschallenging
The lines y=2x+1y = 2x + 1 and y=2x−3y = 2x - 3 are drawn on the same axes. Describe what the graphs show about the simultaneous equations y=2x+1y = 2x + 1 and y=2x−3y = 2x - 3.
Show worked solution

Worked solution

  1. Write down the two equations

    (1)  y=2x+1(2)  y=2x−3(1)\ \ y = 2x + 1 \qquad (2)\ \ y = 2x - 3

    Both are already in the form y = mx + c, which makes the gradients easy to compare.

  2. Read off the gradients

    m1=2,m2=2m_1 = 2, \qquad m_2 = 2

    The number in front of x is the gradient. Both are 2.

  3. Read off the y-intercepts

    c1=1,c2=−3c_1 = 1, \qquad c_2 = -3

    The constant is the y-intercept. One line crosses the y-axis at 1, the other at -3.

  4. Compare

    m1=m2 but c1≠c2m_1 = m_2 \text{ but } c_1 \ne c_2

    Equal gradients mean the lines are equally steep; different intercepts mean they are in different places.

  5. Sketch the graphs

    y=2x+1abovey=2x−3y = 2x + 1 \quad \text{above} \quad y = 2x - 3

    The second line is the first one slid down 4 units. The gap between them is always the same.

  6. Try to solve algebraically

    2x+1=2x−32x + 1 = 2x - 3

    If the lines did cross, the two expressions for y would be equal there.

  7. Subtract 2x from both sides

    1=−31 = -3

    The x-terms cancel completely, leaving a statement that is false.

  8. Interpret the contradiction

    1=−3 is impossible1 = -3 \text{ is impossible}

    No value of x can make this true, so there is no x at which the lines meet.

  9. Test a value of x

    x=0: y=1 and y=−3x = 0: \ y = 1 \text{ and } y = -3

    At x=0x = 0 the two lines are 4 apart.

  10. Test another value of x

    x=5: y=11 and y=7x = 5: \ y = 11 \text{ and } y = 7

    At x=5x = 5 they are still exactly 4 apart — the gap never closes.

  11. Describe the vertical gap

    (2x+1)−(2x−3)=4(2x + 1) - (2x - 3) = 4

    For every x the difference between the two y-values is 4, a constant.

  12. Name the relationship

    The lines are parallel\text{The lines are parallel}

    Two distinct lines with the same gradient are parallel.

  13. Count the crossing points

    Number of intersections=0\text{Number of intersections} = 0

    Parallel lines never meet, so there is no point lying on both.

  14. Link back to the equations

    No point satisfies both equations\text{No point satisfies both equations}

    A solution would have to be a point on both lines, and no such point exists.

  15. Conclude

    No solution\text{No solution}

    The simultaneous equations y=2x+1y = 2x + 1 and y=2x−3y = 2x - 3 have no solution, because their graphs are parallel lines.

Answer
The lines are parallel, so the equations have no solution\text{The lines are parallel, so the equations have no solution}
Question 3
5 markschallenging
The line y=x−1y = x - 1 crosses the line 2x+3y=122x + 3y = 12 at the point PP. Work out the coordinates of PP.
Show worked solution

Worked solution

  1. Decide on a method

    (1)  y=x−1(2)  2x+3y=12(1)\ \ y = x - 1 \qquad (2)\ \ 2x + 3y = 12

    Equation (1) already has y on its own, so substitution is quicker than elimination here.

  2. Label the two equations

    (1)  y=x−1(2)  2x+3y=12(1)\ \ y = x - 1 \qquad (2)\ \ 2x + 3y = 12

    Equation (1) already gives one letter in terms of the other, which is the signal to use substitution rather than elimination.

  3. Spot that (1) gives y in terms of x

    y=x−1y = x - 1

    Anywhere you see y you may write x - 1 instead — they are equal.

  4. Substitute for y in equation (2)

    2x+3(x−1)=122x + 3(x - 1) = 12

    Replace y with the bracket. Equation (2) now contains only x.

  5. Expand the bracket

    2x+3x−3=122x + 3x - 3 = 12

    Multiply each term inside the bracket by 3.

  6. Collect the x-terms

    5x−3=125x - 3 = 12

    Add the two x-terms together.

  7. Move the number to the other side

    5x=12−−3=155x = 12 - -3 = 15

    Undo the addition or subtraction so the x-term is on its own.

  8. Solve for x

    x=155=3x = \frac{15}{5} = 3

    Divide both sides by 5.

  9. Substitute x=3x = 3 into equation (1)

    y=(3)−1y = (3) - 1

    Equation (1) is the quickest way to get y once you know x.

  10. Work out the value

    y=3−1y = 3 - 1

    Do the multiplication first, then the addition or subtraction.

  11. State y

    y=2y = 2

    That gives y=2y = 2.

  12. Check in equation (2)

    2(3)+3(2)=6+6=12 ✓2(3) + 3(2) = 6 + 6 = 12 \ \checkmark

    Equation (2) was rearranged during the working, so checking the pair in the original version confirms nothing went wrong.

  13. State the solution

    x=3, y=2x = 3,\ y = 2

    Both equations are satisfied, so this is the point where the two lines cross.

  14. Check in equation (1)

    (3)−1=2 ✓(3) - 1 = 2 \ \checkmark

    The pair also satisfies the equation that was substituted, so both equations hold.

  15. Interpret the answer graphically

    (3, 2)(3,\ 2)

    The line y=x−1y = x - 1 and the line 2x+3y=122x + 3y = 12 cross at exactly one point, (3, 2).

Answer
(3, 2)(3,\ 2)
Question 4
5 markschallenging
Line L1L_1 passes through (0,5)(0, 5) and (5,0)(5, 0). Line L2L_2 has gradient 33 and passes through (1,0)(1, 0). Work out the coordinates of the point where L1L_1 and L2L_2 cross.
Show worked solution

Worked solution

  1. Find the equation of line L1

    m=0−55−0=−1⇒y=−x+5m = \frac{0 - 5}{5 - 0} = -1 \Rightarrow y = -x + 5

    L1 crosses the y-axis at (0, 5), so c=5c = 5, and the gradient from (0, 5) to (5, 0) is -1.

  2. Find the equation of line L2

    y=3x+c,  0=3(1)+c⇒c=−3y = 3x + c,\ \ 0 = 3(1) + c \Rightarrow c = -3

    L2 has gradient 3, so y=3x+cy = 3x + c. It passes through (1, 0), which gives c=−3c = -3.

  3. Write both equations

    (1)  y=−x+5(2)  y=3x−3(1)\ \ y = -x + 5 \qquad (2)\ \ y = 3x - 3

    Both lines are now in the form y = mx + c, ready to be equated.

  4. Label the two equations

    (1)  y=−x+5(2)  y=3x−3(1)\ \ y = -x + 5 \qquad (2)\ \ y = 3x - 3

    Both equations are already written with y on its own.

  5. Both right-hand sides equal y

    y=−x+5andy=3x−3y = -x + 5 \quad\text{and}\quad y = 3x - 3

    At the point where the graphs cross, the two lines have the same x-value and the same y-value, so the two expressions for y must be equal there.

  6. Set the two expressions equal

    −x+5=3x−3-x + 5 = 3x - 3

    This gives a single equation in x only — solve it to find the x-coordinate of the crossing point.

  7. Collect the x-terms on one side

    −4x+5=−3-4x + 5 = -3

    Subtract 3x from both sides.

  8. Collect the numbers on the other side

    −4x=−3−5-4x = -3 - 5

    Move the constant across by doing the opposite operation.

  9. Simplify

    −4x=−8-4x = -8

    Now there is a simple one-step equation left.

  10. Solve for x

    x=−8−4=2x = \frac{-8}{-4} = 2

    Divide both sides by -4.

  11. Substitute x=2x = 2 into equation (1)

    y=−(2)+5y = -(2) + 5

    Use either equation — (1) is as good as (2).

  12. Work out y

    y=3y = 3

    The y-coordinate of the crossing point is 3.

  13. Check in equation (2)

    3(2)−3=3 ✓3(2) - 3 = 3 \ \checkmark

    Equation (2) gives the same y-value, so the point really does lie on both lines.

  14. State the solution

    x=2, y=3x = 2,\ y = 3

    The graphs cross at (2, 3), and those coordinates are the solution of the simultaneous equations.

  15. Write the crossing point as coordinates

    (2, 3)(2,\ 3)

    The solution of a pair of simultaneous linear equations is exactly the point of intersection of their graphs.

Answer
(2, 3)(2,\ 3)
Question 5
6 markschallenging
A straight line passes through the points (2,7)(2, 7) and (−1,−2)(-1, -2). Find its equation, and hence find the coordinates of the point where it crosses the line y=x+1y = x + 1.
Show worked solution

Worked solution

  1. Find the gradient of the line through the two points

    m=7−(−2)2−(−1)=93=3m = \frac{7 - (-2)}{2 - (-1)} = \frac{9}{3} = 3

    Gradient is the change in y divided by the change in x. Keep the points in the same order top and bottom.

  2. Find the y-intercept

    7=3(2)+c⇒c=17 = 3(2) + c \Rightarrow c = 1

    Substitute one of the points into y=3x+cy = 3x + c to find c.

  3. Write the equation of the first line

    y=3x+1y = 3x + 1

    The line through (2, 7) and (-1, -2) is y=3x+1y = 3x + 1. Check the other point: 3(−1)+1=−23(-1) + 1 = -2, which matches.

  4. Label the two equations

    (1)  y=3x+1(2)  y=x+1(1)\ \ y = 3x + 1 \qquad (2)\ \ y = x + 1

    Both equations are already written with y on its own.

  5. Both right-hand sides equal y

    y=3x+1andy=x+1y = 3x + 1 \quad\text{and}\quad y = x + 1

    At the point where the graphs cross, the two lines have the same x-value and the same y-value, so the two expressions for y must be equal there.

  6. Set the two expressions equal

    3x+1=x+13x + 1 = x + 1

    This gives a single equation in x only — solve it to find the x-coordinate of the crossing point.

  7. Collect the x-terms on one side

    2x+1=12x + 1 = 1

    Subtract x from both sides.

  8. Collect the numbers on the other side

    2x=1−12x = 1 - 1

    Move the constant across by doing the opposite operation.

  9. Simplify

    2x=02x = 0

    Now there is a simple one-step equation left.

  10. Solve for x

    x=02=0x = \frac{0}{2} = 0

    Divide both sides by 2.

  11. Substitute x=0x = 0 into equation (1)

    y=3(0)+1y = 3(0) + 1

    Use either equation — (1) is as good as (2).

  12. Work out y

    y=1y = 1

    The y-coordinate of the crossing point is 1.

  13. Check in equation (2)

    (0)+1=1 ✓(0) + 1 = 1 \ \checkmark

    Equation (2) gives the same y-value, so the point really does lie on both lines.

  14. State the solution

    x=0, y=1x = 0,\ y = 1

    The graphs cross at (0, 1), and those coordinates are the solution of the simultaneous equations.

  15. Write the crossing point as coordinates

    (0, 1)(0,\ 1)

    The solution of a pair of simultaneous linear equations is exactly the point of intersection of their graphs.

Answer
(0, 1)(0,\ 1)

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