Question

solve the following system of equations graphically on the set of axes below.
$y = x + 4$
$y = -dfrac{1}{2}x + 7$
plot two lines by clicking the graph.
click a line to delete it.

Explanation:

Step1: Analyze the first equation \( y = x + 4 \)

This is a linear equation in slope - intercept form \( y=mx + b \), where the slope \( m = 1 \) and the y - intercept \( b = 4 \). We already have some points plotted for this line. To find another point, we can use the slope. From the y - intercept \( (0,4) \), since the slope is 1 (rise over run = 1/1), we can go up 1 unit and right 1 unit to get the point \( (1,5) \), which is also plotted.

Step2: Analyze the second equation \( y=-\frac{1}{2}x + 7 \)

This is also in slope - intercept form with slope \( m=-\frac{1}{2} \) and y - intercept \( b = 7 \). Let's find two points on this line. When \( x = 0 \), \( y=-\frac{1}{2}(0)+7=7 \), so the y - intercept is \( (0,7) \). When \( x = 2 \), \( y=-\frac{1}{2}(2)+7=- 1 + 7 = 6 \), so the point is \( (2,6) \). When \( x = 4 \), \( y=-\frac{1}{2}(4)+7=-2 + 7 = 5 \), so the point is \( (4,5) \). When \( x = 6 \), \( y=-\frac{1}{2}(6)+7=-3 + 7 = 4 \), so the point is \( (6,4) \).

Step3: Find the intersection point

The solution to the system of linear equations graphically is the point where the two lines intersect. We can also solve the system algebraically to verify. Set \( x + 4=-\frac{1}{2}x+7 \).
Add \( \frac{1}{2}x \) to both sides: \( x+\frac{1}{2}x + 4=-\frac{1}{2}x+\frac{1}{2}x+7 \), which simplifies to \( \frac{3}{2}x+4 = 7 \).
Subtract 4 from both sides: \( \frac{3}{2}x+4 - 4=7 - 4 \), so \( \frac{3}{2}x=3 \).
Multiply both sides by \( \frac{2}{3} \): \( x = 3\times\frac{2}{3}=2 \).
Substitute \( x = 2 \) into the first equation \( y=x + 4 \), then \( y=2 + 4=6 \). So the intersection point is \( (2,6) \). When we plot the second line \( y =-\frac{1}{2}x + 7 \), we will see that the two lines intersect at \( (2,6) \).

Answer:

The solution to the system of equations is \( x = 2 \), \( y = 6 \) or the point \( (2,6) \)