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of equations is given. \\begin{cases} y = 0.5x - 10 \\\\ y = (x - 2)(x …

Question

of equations is given.
\

$$\begin{cases} y = 0.5x - 10 \\\\ y = (x - 2)(x + 6) \\end{cases}$$

solution to the system? round the coordinates to the nearest h

Explanation:

Step1: Set equations equal

Since both equal \( y \), set \( 0.5x - 10=(x - 2)(x + 6) \).

Step2: Expand right side

Expand \( (x - 2)(x + 6) \) to \( x^2 + 4x - 12 \).

Step3: Form quadratic equation

Subtract \( 0.5x - 10 \) from both sides: \( x^2 + 4x - 12 - 0.5x + 10 = 0 \), simplify to \( x^2 + 3.5x - 2 = 0 \).

Step4: Use quadratic formula

For \( ax^2 + bx + c = 0 \), \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = 3.5 \), \( c = -2 \).
Calculate discriminant: \( b^2 - 4ac=(3.5)^2 - 4(1)(-2)=12.25 + 8 = 20.25 \).
Then \( x=\frac{-3.5\pm\sqrt{20.25}}{2}=\frac{-3.5\pm4.5}{2} \).

Step5: Solve for x

Two solutions:
\( x_1=\frac{-3.5 + 4.5}{2}=\frac{1}{2}=0.5 \)
\( x_2=\frac{-3.5 - 4.5}{2}=\frac{-8}{2}=-4 \)

Step6: Find y for each x

For \( x = 0.5 \): \( y = 0.5(0.5)-10 = 0.25 - 10=-9.75 \)
For \( x = -4 \): \( y = 0.5(-4)-10=-2 - 10=-12 \)

Answer:

The solutions are \((0.5, -9.75)\) and \((-4, -12)\)