Open Practice Tests

Nonlinear systems — Topic practice

5 exercises · 1 automatically checked · 4 written self-checks · Untimed

Instructions and review guidance

1 question · Untimed practice

Work through the set before checking solutions. Guide question labels match the source lesson references. 4 written responses appear below the question bank, with model answers for self-review; these are not automatically graded.

Guide question labels stay the same when the order changes. Check answers when you are ready to review.

Written self-checks are below the question bank and are excluded from the automatic score.

Written responses are self-review exercises.

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The drill

Question 4

  • Guide question E7.4

For what value of k do y = x2 and y = 4x + k intersect exactly once?

Enter an integer, decimal or fraction without units. Use up to 5 characters, or 6 including a leading minus.

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Written self-checks

Write your complete response before opening the model answer. Compare both your answer and your reasoning.

Guide question E7.1

Find all solutions of y = x2 and y = x + 6.

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(3, 9) and (−2, 4)

Set the two expressions for y equal:

x2 = x + 6 (x − 3)(x + 2) = 0.

The inputs are 3 and −2. Substitute each into y = x2 to get 9 and 4, respectively. Both ordered pairs also satisfy y = x + 6. A system solution is an ordered pair, not merely an x-value. Review 7.01.

Guide question E7.2

Find the solution of y = (x − 2)2 + 3 and y = 3.

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(2, 3) Substitute y = 3 into y = (x − 2)2 + 3:

3 = (x − 2)2 + 3 (x − 2)2 = 0.

Thus x = 2 and y = 3. A repeated root gives one distinct intersection, at the vertex. Review 7.03.

Guide question E7.3

Find all ordered pairs satisfying x + y = 9 and xy = 20.

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(4, 5) and (5, 4)

From x + y = 9, let y = 9 − x. Then

x(9 − x) = 20 x2 − 9x + 20 = (x − 4)(x − 5) = 0.

If x = 4, then y = 5; if x = 5, then y = 4. Both ordered pairs must be retained. Review 7.05.

Guide question E7.5

Find all real solutions of y = x + 12 and y = x.

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(4, 4)

Equating the rules gives x + 12 = x, so x ≥ 0. Squaring leads to

x + 12 = x2 (x − 4)(x + 3) = 0.

Reject x = −3 because the original square root cannot equal a negative output. For x = 4, 16 = 4, so (4, 4) is valid. Review 7.10.