Systems of Equations With a Missing Constant

The key idea

Two lines have no solution when they are parallel (same slope, different intercept) and infinitely many when they are the same line. In coefficient form, the equations'

x

y

- and constant terms must be in the same ratio for infinitely many solutions; only the

x

and

y

ratios match for none.

The trap

Solving $k^{2} =$ something and keeping both signs. Check each value against the constants' ratio; one sign often gives parallel lines instead.

Can Desmos do it?

A slider for $k$ lets you watch the lines turn parallel, but reading an exact value off a slider is slow and risky. Matching ratios takes one line of algebra.

College Board skill: Systems of two linear equations in two variables · Algebra

Try one at each level

One question from each level. The drill has 12 at every level, and moves you up after four right in a row.

Level 1 · Warm-up

In the system below,

a

is a constant.

a x + 7 y = 15

21 x + 21 y = 12

For what value of

a

does the system have no solution?

Show the answer

Answer: 7No solution means the lines are parallel: the

x

- and

y

-coefficients are in the same ratio, but the constants are not. The

y

-coefficients are in the ratio

7 : 21 = 1 : 3

, so

a = \frac{21}{3} = 7

. The constants are

15

and

12

, not in the ratio

1 : 3

, so the lines don't coincide.

Level 2 · Test level

In the system below,

k

is a constant.

k x + 4 y = 24

9 x + k y = 36

For what value of

k

does the system have infinitely many solutions?

Show the answer

Answer: 6Infinitely many solutions needs all three ratios equal:

\frac{k}{9} = \frac{4}{k} = \frac{24}{36}

. From the first two,

k^{2} = 36

, so

k = \pm 6

. The constants' ratio is

\frac{24}{36} = \frac{2}{3}

, which matches

\frac{k}{9}

only when

k = 6

. With

k = - 6

, the lines are parallel and there is no solution.

Level 3 · Beyond the test

In the system below,

a

and

b

are constants and

a > 0

a x + 2 y = 25

15 x + (a + 1) y = b

If the system has infinitely many solutions, what is the value of

a + b

Show the answer

Answer: 80The equations must be multiples of each other:

\frac{a}{15} = \frac{2}{a + 1} = \frac{25}{b}

. From the first two,

a (a + 1) = 30

, so

a = 5

(the other root is negative). Then

\frac{25}{b} = \frac{5}{15}

, so

b = 75

, and

a + b = 80

Practice all 36 questions

Answer in the drill to get instant feedback, climb the levels, and feed your misses into Tailored practice. Free, no account.

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