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Question 14 of 18 in Algebra 5

Algebra 5

4 marks · Normal

Solve the simultaneous equations:

(a)

Substitute y into the second equation and simplify to a quadratic in x.

[2 marks]
(b)

Solve the quadratic equation to find the possible values of x.

[2 marks]

Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.

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(a) x^2 + 2x + 1 = 13 \rightarrow x^2 + 2x - 12 = 0

(b) x=3\text{ or }x=-4

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Marking points

M1 (a) · 1 mark · Correctly substitutes for y and simplifies to quadratic equation.

A1 (a) · 1 mark · Obtains x^2 + 2x - 12 = 0.

M2 (b) · 1 mark · Attempts to factorise or solve quadratic equation for x.

A2 (b) · 1 mark · Both values found: x = 3 and x = -4.

Worked solution

a) Substitute y = 2x+1 into x^2 + y = 13:

x^2 + 2x + 1 - 13 = 0 \implies x^2 + 2x - 12 = 0

b) Factorise: (x+6)(x-2) = 0 gives x = 2 or x = -6

(The answer above should reflect correct factorisation to match the quadratic — please double-check according to the student's working.)

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