Simultaneous equations (linear + circle)

Algebra

(1MA1) · Higher

ORIGINAL

One linear, one quadratic — substitute, solve, and don't forget to find BOTH coordinates.

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Question

Solve the simultaneous equations

y = x + 3

and

x^2 + y^2 = 29

Every step worked, with the reasoning.

1
$x^2 + (x + 3)^2 = 29$
Substitute $y = x + 3$ into the second equation so it's in $x$ only.
2
$x^2 + x^2 + 6x + 9 = 29$
Expand $(x + 3)^2 = x^2 + 6x + 9$ .
3
$2x^2 + 6x - 20 = 0$
Collect terms and move 29 across.
4
$x^2 + 3x - 10 = 0$
Divide through by 2 to simplify.
5
$(x + 5)(x - 2) = 0$
Factorise the quadratic.
6
$x = -5$ or $x = 2$
Solve for $x$ .
7
$x = 2 \Rightarrow y = 5$ ; $x = -5 \Rightarrow y = -2$
Back-substitute each $x$ into $y = x + 3$ to get its $y$ .

Answer: $(2,\ 5)$ and $(-5,\ -2)$