Question

1. solve the system of equations by graphing.

$y = -x + 7$
$5x + 6y = 10$
graph grid

1. blank
2. solve the system of equations by graphing.

$y = \frac{3}{2}x - 2$
$y = 2x - 1$
graph grid

Explanation:

Problem 6: Solve the system \( y = -x + 7 \) and \( 5x + 6y = 10 \) by graphing

Step 1: Rewrite the second equation in slope - intercept form (\(y=mx + b\))

We have the equation \(5x+6y = 10\). Subtract \(5x\) from both sides: \(6y=-5x + 10\). Then divide each term by 6: \(y=-\frac{5}{6}x+\frac{10}{6}=-\frac{5}{6}x+\frac{5}{3}\)

Step 2: Find two points for each line

• For \(y=-x + 7\):

When \(x = 0\), \(y=7\) (so the point is \((0,7)\)). When \(y = 0\), \(0=-x + 7\), so \(x = 7\) (the point is \((7,0)\))

• For \(y=-\frac{5}{6}x+\frac{5}{3}\):

When \(x = 0\), \(y=\frac{5}{3}\approx1.67\) (point \((0,\frac{5}{3})\)). When \(x = 2\), \(y=-\frac{5}{6}(2)+\frac{5}{3}=-\frac{5}{3}+\frac{5}{3}=0\) (point \((2,0)\))

Step 3: Graph the lines and find the intersection

Plot the points for both lines on the coordinate plane and draw the lines. The intersection point of \(y=-x + 7\) and \(y =-\frac{5}{6}x+\frac{5}{3}\) can be found by solving the system algebraically (to check) or by looking at the graph.
Set \(-x + 7=-\frac{5}{6}x+\frac{5}{3}\)
Multiply through by 6 to clear the fractions: \(-6x + 42=-5x+10\)
Add \(6x\) to both sides: \(42=x + 10\)
Subtract 10 from both sides: \(x = 32\)
Then \(y=-32 + 7=-25\)? Wait, that can't be right. Wait, I made a mistake in the algebra. Let's redo the solution of the system:

We have \(y=-x + 7\) and \(5x+6y = 10\)

Substitute \(y=-x + 7\) into \(5x + 6y=10\):

\(5x+6(-x + 7)=10\)

\(5x-6x + 42=10\)

\(-x=10 - 42=-32\)

\(x = 32\), then \(y=-32 + 7=-25\). But this seems odd for a graph with a grid that probably has smaller values. Wait, maybe I misread the equation. Wait, the original equation: is it \(5x + 6y=10\) or maybe \(5x+6y = 42\)? Wait, no, the user provided the problem as is. Alternatively, maybe there is a typo, but assuming the equations are correct, the intersection point is \((32,-25)\). But if we graph the lines:

The line \(y=-x + 7\) has a slope of - 1 and y - intercept 7. The line \(y=-\frac{5}{6}x+\frac{5}{3}\) has a slope of \(-\frac{5}{6}\) and y - intercept \(\frac{5}{3}\). The two lines will intersect at \((32,-25)\)

Problem 8: Solve the system \(y=\frac{3}{2}x-2\) and \(y = 2x-1\) by graphing

Step 1: Find two points for each line

• For \(y=\frac{3}{2}x-2\):

When \(x = 0\), \(y=-2\) (point \((0,-2)\)). When \(x = 2\), \(y=\frac{3}{2}(2)-2=3 - 2 = 1\) (point \((2,1)\))

• For \(y=2x-1\):

When \(x = 0\), \(y=-1\) (point \((0,-1)\)). When \(x = 1\), \(y=2(1)-1 = 1\) (point \((1,1)\))

Step 2: Graph the lines and find the intersection

Plot the points for both lines. The line \(y=\frac{3}{2}x-2\) has a slope of \(\frac{3}{2}\) and y - intercept - 2. The line \(y = 2x-1\) has a slope of 2 and y - intercept - 1.
To find the intersection algebraically, set \(\frac{3}{2}x-2=2x-1\)
Subtract \(\frac{3}{2}x\) from both sides: \(-2=\frac{1}{2}x-1\)
Add 1 to both sides: \(-1=\frac{1}{2}x\)
Multiply both sides by 2: \(x=-2\)
Then \(y=2(-2)-1=-5\)

Answer:

s:

• For problem 6: The solution of the system \(

$$\begin{cases}y=-x + 7\\5x + 6y=10\end{cases}$$

\) is \(x = 32,y=-25\) (after correcting the algebraic solution, but there might be a mis - typing in the original problem as the numbers seem large for a typical graphing problem)

• For problem 8: The solution of the system \(

$$\begin{cases}y=\frac{3}{2}x-2\\y = 2x-1\end{cases}$$

\) is \(x=-2,y=-5\)