<em><u>The solution is (4, 4)</u></em>
<em><u>Solution:</u></em>
<em><u>Given system of equations are:</u></em>

<em><u>Substitute eqn 2 in eqn 1</u></em>

Make the right side of equation 0

<em><u>Solve by quadratic equation</u></em>

<em><u>Substitute x = 4 in eqn 2</u></em>
y = 2(4) - 4
y = 8 - 4
y = 4
Thus solution is (4, 4)
5x - 2y = -6 ⇒ 10x - 4y = -12
2x - 1y = 1 ⇒ <u>10x - 5y = 5</u>
y = -17
5x - 2(-17) = -6
5x + 34 = -6
<u> - 34 - 34</u>
<u>5x</u> = <u>-40</u>
5 5
x = -8
(x, y) = (-8, -17)
2x + 3y = 432 ⇒ 10x + 15y = 2160
5x + 2y = 16 ⇒ <u>10x + 4y = 32</u>
<u>11y</u> = <u>2128</u>
11 11
y = 193.4545455
2x + 3(193.4545455) = 432
2x + 580.3636364 = 432
<u> - 580.3636364 - 580.3636364</u>
<u>2x</u> = <u>-148.3636364</u>
2 2
x = -74.1818182
(x, y) = (-74.1818184, 193.4545455)
Domain: {-4, -2, 0, 2, 4}
range: {-2, -1, 0, 1, 2}
Begin by finding the lowest point the quadratic equation can be, the vertex;
x²-1= is just a translation down of the graph x²
vertex; (0, -1) and since the graph of x² would extend to infinity beyond that point, we can say {x| x≥0} for domain and {y| y≥-1}.
For the linear equation, it is possible to have all x and y values, therefore range and domain belong to all real numbers.
Hope I helped :)