Questions: Solve the system graphically. (If there is no solution, enter NO SOLUTION.) x+y=10 x^2+y^2-10x=0 (x, y)=( )(smaller x-value) (x, y)=( )(larger x-value)

Solution Steps

To solve the system graphically, we need to plot the two equations and find their intersection points. The first equation is a linear equation, and the second equation is a circle. The intersection points of these two graphs will give us the solutions.

First, we rewrite the given system of equations for clarity: \[ \begin{cases} x + y = 10 \\ x^2 + y^2 - 10x = 0 \end{cases} \]

From the first equation, solve for \( y \): \[ y = 10 - x \]

Substitute \( y = 10 - x \) into the second equation: \[ x^2 + (10 - x)^2 - 10x = 0 \]

Expand and simplify the equation: \[ x^2 + (10 - x)^2 - 10x = 0 \] \[ x^2 + (100 - 20x + x^2) - 10x = 0 \] \[ 2x^2 - 30x + 100 = 0 \]

Solve the quadratic equation \( 2x^2 - 30x + 100 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \): \[ a = 2, \quad b = -30, \quad c = 100 \] \[ x = \frac{30 \pm \sqrt{(-30)^2 - 4 \cdot 2 \cdot 100}}{2 \cdot 2} \] \[ x = \frac{30 \pm \sqrt{900 - 800}}{4} \] \[ x = \frac{30 \pm \sqrt{100}}{4} \] \[ x = \frac{30 \pm 10}{4} \] \[ x = \frac{40}{4} = 10 \quad \text{or} \quad x = \frac{20}{4} = 5 \]

For \( x = 10 \): \[ y = 10 - 10 = 0 \] For \( x = 5 \): \[ y = 10 - 5 = 5 \]

Final Answer