Question

Solve the system y=x²-2 and x²+y=4.Illustrate graphically.

Answer 1

Answer:

Solution: $(-\sqrt{3},1),(\sqrt{3},1)$

Step-by-step explanation:

System of equation:

$y=x^2-2\text{ and }x^2+y=4$

Using substitution method.

$y=x^2-2$ substitute into $x^2+y=4$

$x^2+x^2-2=4$              (combine like term)

$2x^2=4+2$                   (Addition property of equality)

$2x^2=6$                       (division property of equality)

$x^2=3$                          (Taking square root)

$x=\pm \sqrt{3}$

Substitute x into  $y=x^2-2$

$y=(\sqrt{3})^2-2$

$y=1$

Solution: $(-\sqrt{3},1),(\sqrt{3},1)$

Now cross check the solution using graph.

Please find attachment for graph.