Choosing a Coefficient in a Pair of Equations

Match the letters first, then look at the numbers.

An unknown coefficient

Here is a pair of equations with a letter k in place of one coefficient: 2x + y = 5 and kx + 2y = 12. The question is which value of k gives no solution, and which values give one.

Match the letters first. The second equation has 2y, so double every term of the first: 4x + 2y = 10. This does not depend on k at all.

The value that gives no solution

If k = 4, the second equation is 4x + 2y = 12. Its letter terms are 4x + 2y, exactly the letter terms of the doubled first equation. So 4x + 2y would have to equal 10 and also equal 12, which is impossible. At k = 4 the pair has no solution.

On a graph, the two lines are equally steep and at different heights, so they are parallel and never meet.

xy4x + 2y = 12

The gold line is 2x + y = 5. At k = 4 the two lines are equally steep and never meet.

When the numbers match too

Suppose instead the second equation were kx + 2y = 10. Then k = 4 gives 4x + 2y = 10, which is the doubled first equation exactly. The two equations draw the same line, and every point on it solves both: infinitely many solutions.

So the choice of k decides whether the letter terms match, and then the numbers on the right decide between no solution and infinitely many.

Every other value gives one crossing

For any k other than 4, the letter terms cannot be matched, so the lines are not equally steep and they cross exactly once. Take k = 6: the second equation is 6x + 2y = 12. Halve every term: 3x + y = 6.

Subtract 2x + y = 5 from 3x + y = 6: x = 1. Then 2 × 1 + y = 5, so y = 3. Check in the original second equation: 6 × 1 + 2 × 3 = 6 + 6 = 12. The one solution is x = 1 and y = 3.

xy6x + 2y = 12

The gold line is 2x + y = 5 again. At k = 6 the second line is steeper, and it crosses the gold line once, where x = 1 and y = 3.

y = x + 1(0.67, 1.67)y = −0.5x + 2x = 0.67, y = 1.67−4−4−2−22244

m₁ ≠ m₂: the two lines share exactly one point, and its coordinates are the one pair (x, y) that satisfies both equations

Make the gradients equal with different intercepts

Choosing a coefficient chooses how steep a line is. The line y = x + 1 stays fixed. Turn the gold line with one handle and slide it with the other. Only one steepness makes it parallel to y = x + 1; every other steepness makes the lines cross once.

The usual mistakes

Comparing coefficients before matching the letters. In 2x + y = 5 and kx + 2y = 12, answering k = 2 because the first equation has 2x ignores the y terms, which are 1y and 2y. Scale first so that one letter matches, then compare the other.

Saying that matching letter terms always give infinitely many solutions. That is so only when the numbers on the right match as well. Here 10 and 12 differ, so k = 4 gives no solution.

Worked example: An Order Recovered from Two Suppliers' Quotes

Question A grocer ordered x crates of apples and y crates of oranges, and then lost the order sheet. Two suppliers had priced the same order. Supplier P charges $8 a crate of apples and $5 a crate of oranges, and quoted $260. Supplier Q charges $12 a crate of apples and k dollars a crate of oranges, and quoted $404. (a) Write an equation for each quote. Find the value of k for which this pair of equations has no solution, and explain why. (b) Supplier Q in fact charges $8 a crate of oranges. Find the order.

  1. 1.Supplier P's quote gives 8x + 5y = 260 (1). Supplier Q's quote gives 12x + ky = 404 (2).

    8x + 5y=260(1) Supplier P12x + ky=404(2) Supplier Q
    (1) Supplier P8x + 5y=260(2) Supplier Q12x + ky=404
    Each quote is one equation: 8x + 5y = 260 for Supplier P and 12x + ky = 404 for Supplier Q.

  2. 2.Match the x terms: multiply equation (1) by 1.5, which gives 12x + 7.5y = 390 (3).

    8x + 5y=260(1) Supplier P12x + ky=404(2) Supplier Q12x + 7.5y=390(3), which is (1) × 1.5
    (1) Supplier P8x + 5y=260(2) Supplier Q12x + ky=404(3), which is (1) × 1.512x + 7.5y=390
    Multiply equation (1) by 1.5 so that its x term matches: 12x + 7.5y = 390.

  3. 3.If k = 7.5, equations (2) and (3) have the same left side, 12x + 7.5y, but different right sides, 404 and 390. No pair (x, y) can make one expression equal two different numbers, so there is no solution. On a graph the two lines are parallel.

    8x + 5y=260(1) Supplier P12x + ky=404(2) Supplier Q12x + 7.5y=390(3), which is (1) × 1.512x + 7.5y=404(2) with k = 7.5one left side cannot equal both 390 and 404
    (1) Supplier P8x + 5y=260(2) Supplier Q12x + ky=404(3), which is (1) × 1.512x + 7.5y=390(2) with k = 7.512x + 7.5y=404one left side cannot equal both 390 and 404
    If k = 7.5 the left sides match, but one would equal 390 and the other 404. No pair (x, y) does both.

  4. 4.(a) k = 7.5. At that price every price at Supplier Q would be 1.5 times the price at Supplier P, so Q's quote for any order would be 1.5 × 260 = $390, never $404. For any other k the y terms differ and the pair has exactly one solution.

    8x + 5y=260(1) Supplier P12x + ky=404(2) Supplier Q12x + 7.5y=390(3), which is (1) × 1.512x + 7.5y=404(2) with k = 7.5no solution: Q would always charge 1.5 × Pand 1.5 × 260 = 390, never 404
    (1) Supplier P8x + 5y=260(2) Supplier Q12x + ky=404(3), which is (1) × 1.512x + 7.5y=390(2) with k = 7.512x + 7.5y=404no solution: Q would always charge 1.5 × Pand 1.5 × 260 = 390, never 404
    (a) k = 7.5: the pair has no solution, because Q's quote would always be 1.5 times P's.

  5. 5.With k = 8, subtract equation (3) from equation (2): 0.5y = 14, so y = 28. Substitute into (1): 8x + 140 = 260, so 8x = 120 and x = 15.

    12x + 8y=404(2) with k = 812x + 7.5y=390(3)0.5y=14(2) − (3)y=28multiply both sides by 28x + 140=260put y = 28 in (1)x=15subtract 140, then divide by 8
    (2) with k = 812x + 8y=404(3)12x + 7.5y=390(2) − (3)0.5y=14multiply both sides by 2y=28put y = 28 in (1)8x + 140=260subtract 140, then divide by 8x=15
    With k = 8, subtract (3) from (2): 0.5y = 14, so y = 28, and then x = 15.

  6. 6.(b) The order was 15 crates of apples and 28 crates of oranges. Check with Supplier Q: 12 × 15 + 8 × 28 = 180 + 224 = 404.

    15 crates of apples and 28 crates of orangesSupplier P15 × $8 = $12028 × $5 = $140$260Supplier Q15 × $12 = $18028 × $8 = $224$404check in (2): 180 + 224 = 404
    15 crates of apples and 28 crates of orangesSupplier P$120$140$260Supplier Q$180$224$404check in (2): 180 + 224 = 404
    (b) The order was 15 crates of apples and 28 crates of oranges: 180 + 224 = 404.

Answer: (a) 8x + 5y = 260 and 12x + ky = 404; k = 7.5, since Q's quote would then always be 1.5 times P's; (b) 15 crates of apples and 28 crates of oranges

Common mistakes

  • Comparing the y coefficients 5 and k directly and answering k = 5. The letters must be matched first: once the x terms agree, the y coefficient of equation (1) has become 7.5.
  • Saying that k = 7.5 gives infinitely many solutions because the left sides match. That happens only when the right sides match too. Here 390 and 404 differ, so there is no solution.

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