Question

1 solve the system of equations by graphing. equation 1: $y + x = 2$ (a) equation 2: $-8 - 2y + x = 0$

Explanation:

Step1: Rewrite Equation 1 in slope - intercept form

We have the equation \(y + x=2\). To get it in the form \(y = mx + b\) (where \(m\) is the slope and \(b\) is the y - intercept), we solve for \(y\). Subtract \(x\) from both sides: \(y=-x + 2\). The slope \(m_1=- 1\) and the y - intercept \(b_1 = 2\).

Step2: Rewrite Equation 2 in slope - intercept form

We have the equation \(-8-2y + x = 0\). First, we isolate \(y\). Add \(2y\) to both sides: \(x-8=2y\). Then divide both sides by 2: \(y=\frac{1}{2}x-4\). The slope \(m_2=\frac{1}{2}\) and the y - intercept \(b_2=-4\).

Step3: Find two points for each line

For Equation 1 (\(y=-x + 2\)):

• When \(x = 0\), \(y=-0 + 2=2\). So one point is \((0,2)\).
• When \(x = 2\), \(y=-2 + 2=0\). So another point is \((2,0)\).

For Equation 2 (\(y=\frac{1}{2}x-4\)):

• When \(x = 0\), \(y=\frac{1}{2}(0)-4=-4\). So one point is \((0,-4)\).
• When \(x = 8\), \(y=\frac{1}{2}(8)-4=4 - 4=0\). So another point is \((8,0)\).

Step4: Graph the lines and find the intersection

When we graph the line \(y=-x + 2\) (using the points \((0,2)\) and \((2,0)\)) and the line \(y=\frac{1}{2}x-4\) (using the points \((0,-4)\) and \((8,0)\)), we find that the two lines intersect at the point \((4,-2)\). We can also verify this by substituting \(x = 4\) and \(y=-2\) into both equations:

• For Equation 1: \(y+x=-2 + 4=2\), which matches the right - hand side of \(y + x=2\).
• For Equation 2: \(-8-2y+x=-8-2(-2)+4=-8 + 4+4=0\), which matches the right - hand side of \(-8-2y + x=0\).

Answer:

The solution to the system of equations is \(x = 4\), \(y=-2\) (or the ordered pair \((4,-2)\))