Simultaneous Equations Practice Questions and Solutions

Les exercices

1. Exercice 1Base3 points

   Solve x + y = 10 and x - y = 4

   Afficher la correctionMasquer la correction

   x = 7, y = 3

   1. 1The y terms are +y and -y, so adding the two equations removes y.
   2. 2(x + y) + (x - y) = 10 + 4 gives 2x = 14, so x = 7.
   3. 3Substitute into the first equation: 7 + y = 10, so y = 3.
   4. 4Check in the second equation, which was not used: 7 - 3 = 4.
2. Exercice 2Base3 points

   Solve 2x + y = 11 and x - y = 1

   Afficher la correctionMasquer la correction

   x = 4, y = 3

   1. 1The y terms are +y and -y, so add the equations.
   2. 23x = 12, so x = 4.
   3. 3Substitute into the second equation: 4 - y = 1, so y = 3.
   4. 4Check in the first equation: 2(4) + 3 = 8 + 3 = 11.
3. Exercice 3Base3 points

   Solve 3x + 2y = 19 and 3x - y = 4

   Afficher la correctionMasquer la correction

   x = 3, y = 5

   1. 1The x terms are identical at 3x, so subtract the second equation from the first.
   2. 22y - (-y) = 3y and 19 - 4 = 15, so 3y = 15 and y = 5.
   3. 3Substitute into the second equation: 3x - 5 = 4, so 3x = 9 and x = 3.
   4. 4Check in the first equation: 3(3) + 2(5) = 9 + 10 = 19.
4. Exercice 4Standard3 points

   Solve 2x + 3y = 12 and x + y = 5

   Afficher la correctionMasquer la correction

   x = 3, y = 2

   1. 1The second equation rearranges in one step: x = 5 - y.
   2. 2Substitute into the first: 2(5 - y) + 3y = 12.
   3. 310 - 2y + 3y = 12, so 10 + y = 12 and y = 2.
   4. 4Then x = 5 - 2 = 3.
   5. 5Check in the first equation: 2(3) + 3(2) = 6 + 6 = 12.
5. Exercice 5Standard4 points

   Solve 3x + 4y = 25 and 2x + 3y = 18

   Afficher la correctionMasquer la correction

   x = 3, y = 4

   1. 1Nothing matches, so aim at the y terms: the lowest common multiple of 4 and 3 is 12.
   2. 2Multiply the first equation by 3: 9x + 12y = 75. Multiply the second by 4: 8x + 12y = 72.
   3. 3Both y terms are now +12y, so subtract: x = 3.
   4. 4Substitute into 2x + 3y = 18: 6 + 3y = 18, so 3y = 12 and y = 4.
   5. 5Check in 3x + 4y = 25: 9 + 16 = 25.
6. Exercice 6Standard4 points

   Solve 5x - 2y = 16 and 3x + 4y = 20

   Afficher la correctionMasquer la correction

   x = 4, y = 2

   1. 1Only one equation needs multiplying: double the first to give 10x - 4y = 32.
   2. 2The y terms are now -4y and +4y, so add the equations: 13x = 52 and x = 4.
   3. 3Substitute into 3x + 4y = 20: 12 + 4y = 20, so 4y = 8 and y = 2.
   4. 4Check in 5x - 2y = 16: 20 - 4 = 16.
7. Exercice 7Standard3 points

   Solve y = 2x - 1 and 3x + y = 14

   Afficher la correctionMasquer la correction

   x = 3, y = 5

   1. 1The first equation already gives y on its own, so substitute the whole expression into the second.
   2. 23x + (2x - 1) = 14
   3. 35x - 1 = 14, so 5x = 15 and x = 3.
   4. 4Then y = 2(3) - 1 = 5.
   5. 5Check in 3x + y = 14: 9 + 5 = 14.
8. Exercice 8Standard5 points

   Three coffees and two teas cost PKR 1,150. One coffee and four teas cost PKR 950. Find the price of each.

   Afficher la correctionMasquer la correction

   A coffee is PKR 270 and a tea is PKR 170

   1. 1Let c be the price of a coffee and t the price of a tea: 3c + 2t = 1150 and c + 4t = 950.
   2. 2The second equation gives c = 950 - 4t.
   3. 3Substitute into the first: 3(950 - 4t) + 2t = 1150, so 2850 - 12t + 2t = 1150.
   4. 4-10t = -1700, so t = 170.
   5. 5Then c = 950 - 4(170) = 950 - 680 = 270.
   6. 6Check in the first equation: 3(270) + 2(170) = 810 + 340 = 1150.
9. Exercice 9Approfondi5 points

   Solve y = x² - 3x + 4 and y = x + 1

   Afficher la correctionMasquer la correction

   x = 1, y = 2 and x = 3, y = 4

   1. 1Both equations give y, so set the right-hand sides equal: x² - 3x + 4 = x + 1.
   2. 2Rearrange to zero: x² - 4x + 3 = 0.
   3. 3Factorise: (x - 1)(x - 3) = 0, so x = 1 or x = 3.
   4. 4Each x has its own y, taken from y = x + 1: x = 1 gives y = 2, and x = 3 gives y = 4.
   5. 5Check both in the quadratic: 1 - 3 + 4 = 2 and 9 - 9 + 4 = 4.
10. Exercice 10Approfondi6 points

    Solve x² + y² = 25 and y = x + 1

    Afficher la correctionMasquer la correction

    x = -4, y = -3 and x = 3, y = 4

    1. 1Substitute y = x + 1 into the first equation: x² + (x + 1)² = 25.
    2. 2(x + 1)² = x² + 2x + 1, so the equation becomes 2x² + 2x + 1 = 25.
    3. 32x² + 2x - 24 = 0, and dividing every term by 2 gives x² + x - 12 = 0.
    4. 4Factorise: (x + 4)(x - 3) = 0, so x = -4 or x = 3.
    5. 5From y = x + 1: x = -4 gives y = -3, and x = 3 gives y = 4.
    6. 6Check in x² + y² = 25: 16 + 9 = 25 and 9 + 16 = 25.

Which method each pair is asking for

Elimination and substitution both work on every linear pair, but one of them is usually much quicker, and the equations say which. If a variable already has matching coefficients, or coefficients that differ only in sign, eliminate. If one equation already has a variable on its own - y = 2x - 1, or x + y = 5, which rearranges in one step - substitute.

Where nothing matches, elimination still wins, but one or both equations have to be multiplied first. The target is the lowest common multiple of the two coefficients: for 4y and 3y that is 12y, so multiply by 3 and by 4 respectively. Choosing the variable with the smaller multiple keeps the numbers manageable.

• Coefficients match or differ only in sign - eliminate, no multiplying
• One equation gives a variable on its own - substitute
• Nothing matches - multiply to the lowest common multiple, then eliminate
• One equation is quadratic - substitute, always

Check in the equation you did not use

Once the first variable is found, the second comes from substituting into one of the two equations. That equation is then useless as a check, because the value was manufactured to fit it. Substituting back into it will agree even when the first variable is wrong.

The check has to happen in the other equation. In every solution below, the final line uses whichever equation was not used at the substitution stage, which is the only one carrying independent information.

When one equation is a curve

The last two questions pair a straight line with a curve - a parabola in one, a circle in the other. Elimination does not work here, because the terms are not alike. Substitution does, and it always produces a quadratic in one variable.

That quadratic normally has two solutions, and geometrically that is the line cutting the curve at two points. So the answer is two coordinate pairs, not two values of x. Every x has to be taken back to the linear equation to find its own y. On Pearson Edexcel GCSE Mathematics (1MA1), this linear-and-quadratic case is Higher tier content; the purely linear pairs above it appear at both tiers.