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رياضيات GCSE

Simultaneous Equations Practice Questions and Solutions

Ten pairs solved by elimination and substitution, including a line meeting a curve, each checked in the equation that was not used to find the second value.

الإجابة باختصار

Ten pairs of simultaneous equations worked in full: three by straight elimination, two where one or both equations must be multiplied first, two by substitution, one worded, and two where a straight line meets a curve. Every solution is checked in the equation not used to find the second value.

الأسئلة

  1. سؤال 1أساسي3 درجات

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

    أظهر الحل

    x = 7, y = 3

    1. The y terms are +y and -y, so adding the two equations removes y.
    2. (x + y) + (x - y) = 10 + 4 gives 2x = 14, so x = 7.
    3. Substitute into the first equation: 7 + y = 10, so y = 3.
    4. Check in the second equation, which was not used: 7 - 3 = 4.
  2. سؤال 2أساسي3 درجات

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

    أظهر الحل

    x = 4, y = 3

    1. The y terms are +y and -y, so add the equations.
    2. 3x = 12, so x = 4.
    3. Substitute into the second equation: 4 - y = 1, so y = 3.
    4. Check in the first equation: 2(4) + 3 = 8 + 3 = 11.
  3. سؤال 3أساسي3 درجات

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

    أظهر الحل

    x = 3, y = 5

    1. The x terms are identical at 3x, so subtract the second equation from the first.
    2. 2y - (-y) = 3y and 19 - 4 = 15, so 3y = 15 and y = 5.
    3. Substitute into the second equation: 3x - 5 = 4, so 3x = 9 and x = 3.
    4. Check in the first equation: 3(3) + 2(5) = 9 + 10 = 19.
  4. سؤال 4جوهري3 درجات

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

    أظهر الحل

    x = 3, y = 2

    1. The second equation rearranges in one step: x = 5 - y.
    2. Substitute into the first: 2(5 - y) + 3y = 12.
    3. 10 - 2y + 3y = 12, so 10 + y = 12 and y = 2.
    4. Then x = 5 - 2 = 3.
    5. Check in the first equation: 2(3) + 3(2) = 6 + 6 = 12.
  5. سؤال 5جوهري4 درجات

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

    أظهر الحل

    x = 3, y = 4

    1. Nothing matches, so aim at the y terms: the lowest common multiple of 4 and 3 is 12.
    2. Multiply the first equation by 3: 9x + 12y = 75. Multiply the second by 4: 8x + 12y = 72.
    3. Both y terms are now +12y, so subtract: x = 3.
    4. Substitute into 2x + 3y = 18: 6 + 3y = 18, so 3y = 12 and y = 4.
    5. Check in 3x + 4y = 25: 9 + 16 = 25.
  6. سؤال 6جوهري4 درجات

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

    أظهر الحل

    x = 4, y = 2

    1. Only one equation needs multiplying: double the first to give 10x - 4y = 32.
    2. The y terms are now -4y and +4y, so add the equations: 13x = 52 and x = 4.
    3. Substitute into 3x + 4y = 20: 12 + 4y = 20, so 4y = 8 and y = 2.
    4. Check in 5x - 2y = 16: 20 - 4 = 16.
  7. سؤال 7جوهري3 درجات

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

    أظهر الحل

    x = 3, y = 5

    1. The first equation already gives y on its own, so substitute the whole expression into the second.
    2. 3x + (2x - 1) = 14
    3. 5x - 1 = 14, so 5x = 15 and x = 3.
    4. Then y = 2(3) - 1 = 5.
    5. Check in 3x + y = 14: 9 + 5 = 14.
  8. سؤال 8جوهري5 درجات

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

    أظهر الحل

    A coffee is PKR 270 and a tea is PKR 170

    1. Let c be the price of a coffee and t the price of a tea: 3c + 2t = 1150 and c + 4t = 950.
    2. The second equation gives c = 950 - 4t.
    3. Substitute into the first: 3(950 - 4t) + 2t = 1150, so 2850 - 12t + 2t = 1150.
    4. -10t = -1700, so t = 170.
    5. Then c = 950 - 4(170) = 950 - 680 = 270.
    6. Check in the first equation: 3(270) + 2(170) = 810 + 340 = 1150.
  9. سؤال 9متقدّم5 درجات

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

    أظهر الحل

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

    1. Both equations give y, so set the right-hand sides equal: x² - 3x + 4 = x + 1.
    2. Rearrange to zero: x² - 4x + 3 = 0.
    3. Factorise: (x - 1)(x - 3) = 0, so x = 1 or x = 3.
    4. Each x has its own y, taken from y = x + 1: x = 1 gives y = 2, and x = 3 gives y = 4.
    5. Check both in the quadratic: 1 - 3 + 4 = 2 and 9 - 9 + 4 = 4.
  10. سؤال 10متقدّم6 درجات

    Solve x² + y² = 25 and y = x + 1

    أظهر الحل

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

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

أين يقع الخطأ عادةً

  • Adding the equations when the matching terms have the same sign, or subtracting when they have opposite signs, so the variable survives instead of cancelling.
  • Subtracting a negative term carelessly: in 3x + 2y = 19 minus 3x - y = 4, writing 2y - y = y instead of 2y - (-y) = 3y.
  • Finding one value and stopping. The answer to a pair of simultaneous equations is a pair of numbers, and half of it earns very little.
  • Checking in the equation just used to find the second value. It will agree whatever the first value was, so it tests nothing.
  • Giving two x values and a single y on a line-and-curve question. Each x has its own y, and the answer is two coordinate pairs.

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.

أسئلة شائعة

How do I know whether to add or subtract the equations?

Look at the signs of the variable you are eliminating. Same sign, subtract; opposite signs, add. In 5x - 2y = 16 doubled to 10x - 4y = 32, the y terms are -4y and +4y, so adding removes them. Deciding by looking at the signs takes a second and removes the most common slip in the topic.

Do simultaneous equations always have exactly one solution?

Linear pairs usually do, and it is the point where the two lines cross. If the lines are parallel there is no solution, and if the two equations are the same line there are infinitely many. A line and a curve can cross twice, touch once, or miss entirely - which is why the quadratic that comes out can have two roots, one, or none.

Can I solve these on a graph instead?

You can, and the crossing point is the solution, but a graph only gives what you can read off it. If the answer is x = 3, y = 4 the graph will show it clearly; if it is x = 3.4, y = -1.7 it will not. Algebra gives the exact values, and an exam that says solve algebraically will not accept a graph.

Are linear and quadratic pairs on the Foundation paper?

Not on Pearson Edexcel 1MA1. Solving a pair where one equation is linear and one is quadratic is listed as Higher tier content. Foundation candidates meet simultaneous linear equations only, so a Foundation student working through this page can stop after the worded question.

المصادر

  1. Pearson Edexcel GCSE Mathematics (9-1) from 2015 - specification — Pearson Edexcel

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