2.5 Simultaneous equations by elimination and substitution — Practise

Work through these on your own. Decide on the method before you write anything: elimination for two straight lines, substitution when one equation is a curve. Write every solution as a pair and check it in both equations.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Solve the simultaneous equations $2x + 3y = 7$ and $5x - y = 9$ .
Solve $3x + 2y = 16$ and $2x - y = 6$ .
Solve $y = x + 1$ and $x^2 + y^2 = 25$ .
Solve $y = 2x + 1$ and $y = x^2 - 2$ .
Solve $x + y = 5$ and $xy = 6$ .
Solve $y = x^2$ and $y = 2x + 3$ .
Solve $x + y = 4$ and $x^2 - y^2 = 8$ .
Solve $x^2 + y^2 = 4$ and $y = x + 2$ .
Solve $x + y = 5$ and $x^2 + y^2 = 13$ .
Solve $x + y = 1$ and $x^2 + y^2 = 1$ .
Solve $x + 2y = 4$ and $x^2 + y^2 = 16$ .
The curve $y = x^2$ and the line $y = 4$ meet at two points. Solve the pair of equations to find them, and explain what the solutions mean on a graph.

Answers and concise solutions

1. The solution

Answer: Multiply the second equation by $3$ : $15x - 3y = 27$ . Adding removes $y$ : $17x = 34$ , so $x = 2$ . Substitute back: $5 \times 2 - y = 9$ , so $y = 1$ . The solution is $(2, 1)$ .

2. The solution

Answer: Multiply the second equation by $2$ : $4x - 2y = 12$ . Add the two equations to remove $y$ : $7x = 28$ , so $x = 4$ . Then $2 \times 4 - y = 6$ gives $y = 2$ . The solution is $(4, 2)$ .

3. The solutions

Answer: Substitute $y = x + 1$ : $x^2 + (x + 1)^2 = 25$ , so $2x^2 + 2x - 24 = 0$ and $x^2 + x - 12 = 0$ . Factorise to $(x + 4)(x - 3) = 0$ , so $x = 3$ or $x = -4$ . Then $y = x + 1$ gives the pairs $(3, 4)$ and $(-4, -3)$ .

4. The solutions

Answer: Set the right sides equal: $2x + 1 = x^2 - 2$ , so $x^2 - 2x - 3 = 0$ and $(x - 3)(x + 1) = 0$ . So $x = 3$ or $x = -1$ . Substituting into $y = x^2 - 2$ gives $(3, 7)$ and $(-1, -1)$ .

5. The solutions

Answer: From $x + y = 5$ , $y = 5 - x$ . Substitute into $xy = 6$ : $x(5 - x) = 6$ , so $x^2 - 5x + 6 = 0$ and $(x - 2)(x - 3) = 0$ . So $x = 2$ or $x = 3$ , giving the pairs $(2, 3)$ and $(3, 2)$ .

6. The solutions

Answer: Set the expressions equal: $x^2 = 2x + 3$ , so $x^2 - 2x - 3 = 0$ and $(x - 3)(x + 1) = 0$ . So $x = 3$ or $x = -1$ . Use $y = x^2$ to get the pairs $(3, 9)$ and $(-1, 1)$ .

7. The solution

Answer: Factorise the difference of two squares: $x^2 - y^2 = (x + y)(x - y) = 8$ . Since $x + y = 4$ , this gives $4(x - y) = 8$ , so $x - y = 2$ . Adding $x + y = 4$ and $x - y = 2$ gives $2x = 6$ , so $x = 3$ and $y = 1$ . The solution is $(3, 1)$ .

8. The solutions

Answer: Substitute $y = x + 2$ : $x^2 + (x + 2)^2 = 4$ , so $2x^2 + 4x + 4 = 4$ and $2x^2 + 4x = 0$ . Factorise as $2x(x + 2) = 0$ , so $x = 0$ or $x = -2$ . The pairs are $(0, 2)$ and $(-2, 0)$ .

9. The solutions

Answer: From $y = 5 - x$ , substitute: $x^2 + (5 - x)^2 = 13$ , so $2x^2 - 10x + 25 = 13$ and $x^2 - 5x + 6 = 0$ . Factorise to $(x - 2)(x - 3) = 0$ , so $x = 2$ or $x = 3$ , giving $(2, 3)$ and $(3, 2)$ .

10. The solutions

Answer: From $y = 1 - x$ , substitute: $x^2 + (1 - x)^2 = 1$ , so $2x^2 - 2x + 1 = 1$ and $2x(x - 1) = 0$ . So $x = 0$ or $x = 1$ , giving $(0, 1)$ and $(1, 0)$ .

11. The solutions

Answer: From $x = 4 - 2y$ , substitute: $(4 - 2y)^2 + y^2 = 16$ , so $5y^2 - 16y = 0$ . Factorise as $y(5y - 16) = 0$ , so $y = 0$ or $y = \frac{16}{5}$ . Then $x = 4$ or $x = -\frac{12}{5}$ . The pairs are $(4, 0)$ and $\left(-\frac{12}{5}, \frac{16}{5}\right)$ .

12. The points and their meaning

Answer: Substitute $y = 4$ into $y = x^2$ : $x^2 = 4$ , so $x = 2$ or $x = -2$ . The points are $(2, 4)$ and $(-2, 4)$ . On a graph these are the two places where the horizontal line $y = 4$ cuts the parabola, one on each side of the $y$ -axis.