Question

solve the system of equations by graphing. first graph the equations, and then identify the solutions.
( y = \frac{4}{5}x - 2 )
( y = (x - 4)^2 + 2 )
to graph a linear equation, click to plot points on the graph. to graph a quadratic equation, click to plot the vertex first, then plot another point on the graph.

Explanation:

Step1: Analyze the linear equation \( y = \frac{4}{5}x - 2 \)

To graph the linear equation, we can find two points.

• When \( x = 0 \), \( y=\frac{4}{5}(0)-2=-2 \), so one point is \( (0, -2) \).
• When \( x = 5 \), \( y=\frac{4}{5}(5)-2 = 4 - 2=2 \), so another point is \( (5, 2) \).

Step2: Analyze the quadratic equation \( y=(x - 4)^2+2 \)

The quadratic equation is in vertex form \( y = a(x - h)^2 + k \), where the vertex is \( (h,k)=(4,2) \).
To find another point, let's take \( x = 3 \): \( y=(3 - 4)^2+2=1 + 2 = 3 \), so the point is \( (3,3) \).
Take \( x = 5 \): \( y=(5 - 4)^2+2=1 + 2 = 3 \), so the point is \( (5,3) \).

Step3: Find the intersection points

By graphing both equations (plotting the points we found), we can see that the linear equation \( y=\frac{4}{5}x - 2 \) and the quadratic equation \( y=(x - 4)^2+2 \) intersect at two points.

• For the linear equation, when \( x = 5 \), \( y = 2 \). Let's check in the quadratic equation: \( y=(5 - 4)^2+2=3

eq2 \), wait, maybe we made a mistake. Let's solve the system algebraically to confirm.
Set \( \frac{4}{5}x - 2=(x - 4)^2+2 \)
Expand the right side: \( \frac{4}{5}x - 2=x^2-8x + 16+2 \)
Multiply both sides by 5 to eliminate fraction: \( 4x-10 = 5x^2-40x + 90 \)
Rearrange to standard quadratic form: \( 5x^2-44x + 100 = 0 \)
Use quadratic formula \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \), where \( a = 5 \), \( b=-44 \), \( c = 100 \)
Discriminant \( D=(-44)^2-4\times5\times100=1936 - 2000=-64 \). Wait, discriminant is negative? But that contradicts the graphing idea. Wait, maybe calculation error.
Wait, let's re - solve the equation:
\( \frac{4}{5}x-2=(x - 4)^2+2 \)
\( \frac{4}{5}x-2=x^{2}-8x + 16 + 2 \)
\( \frac{4}{5}x-2=x^{2}-8x + 18 \)
Multiply both sides by 5: \( 4x-10 = 5x^{2}-40x + 90 \)
\( 5x^{2}-44x + 100 = 0 \)
Wait, \( b^{2}-4ac=(-44)^{2}-4\times5\times100=1936 - 2000=-64 \). So the system has no real solutions? But that can't be. Wait, maybe the linear equation was misread. Wait, the linear equation is \( y=\frac{4}{5}x-2 \) and quadratic is \( y=(x - 4)^2+2 \). Let's check the value of the linear equation at \( x = 4 \): \( y=\frac{4}{5}\times4-2=\frac{16}{5}-2=\frac{16 - 10}{5}=\frac{6}{5}=1.2 \), and the quadratic at \( x = 4 \) is \( y = 2 \). The linear equation at \( x = 0 \) is \( y=-2 \), quadratic at \( x = 0 \) is \( (0 - 4)^2+2=16 + 2 = 18 \). The linear equation at \( x = 10 \) is \( y=\frac{4}{5}\times10-2=8 - 2 = 6 \), quadratic at \( x = 10 \) is \( (10 - 4)^2+2=36 + 2 = 38 \). So the linear function is below the quadratic function for all real \( x \), so they never intersect? But that's a contradiction to the initial thought. Wait, maybe the linear equation is \( y=\frac{4}{5}x + 2 \)? No, the problem says \( y=\frac{4}{5}x-2 \).
Wait, let's re - check the discriminant calculation. \( a = 5 \), \( b=-44 \), \( c = 100 \). \( D=b^{2}-4ac=(-44)^{2}-4\times5\times100=1936 - 2000=-64 \). Since the discriminant is negative, the system of equations has no real solutions.

Answer:

The system of equations \(

$$\begin{cases}y=\frac{4}{5}x - 2\\y=(x - 4)^2+2\end{cases}$$

\) has no real solutions.