Question

1. $y = 0.5x^2 - 8x + 13$ $y = x - 3$

Explanation:

Step1: Substitute \( y = x - 3 \) into the first equation

Since \( y = x - 3 \), we can replace \( y \) in \( y = 0.5x^{2}-8x + 13 \) with \( x - 3 \). So we get the equation:
\( x - 3=0.5x^{2}-8x + 13 \)

Step2: Rearrange the equation to standard quadratic form

Multiply both sides of the equation by 2 to eliminate the decimal:
\( 2(x - 3)=2\times(0.5x^{2}-8x + 13) \)
\( 2x-6=x^{2}-16x + 26 \)
Then, move all terms to one side to get a quadratic equation in standard form \( ax^{2}+bx + c = 0 \):
\( x^{2}-16x + 26-2x + 6 = 0 \)
\( x^{2}-18x + 32 = 0 \)

Step3: Solve the quadratic equation

We can solve the quadratic equation \( x^{2}-18x + 32 = 0 \) using the quadratic formula \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \), where \( a = 1 \), \( b=-18 \), \( c = 32 \).
First, calculate the discriminant \( \Delta=b^{2}-4ac=(-18)^{2}-4\times1\times32=324 - 128 = 196 \)
Then, \( x=\frac{18\pm\sqrt{196}}{2}=\frac{18\pm14}{2} \)
We have two solutions for \( x \):

• When we take the plus sign: \( x_{1}=\frac{18 + 14}{2}=\frac{32}{2}=16 \)
• When we take the minus sign: \( x_{2}=\frac{18-14}{2}=\frac{4}{2}=2 \)

Step4: Find the corresponding \( y \) values

Now, substitute the \( x \) values back into the equation \( y=x - 3 \) to find the corresponding \( y \) values.

• For \( x = 16 \): \( y=16-3 = 13 \)
• For \( x = 2 \): \( y=2-3=-1 \)

Answer:

The solutions of the system of equations are \( (x = 16,y = 13) \) and \( (x = 2,y=-1) \)