Question

NO LINKS!! Use the method of substitution to solve the system. (If there's no solution, enter no solution). Part 7z​

Answer 1

`(x,y)=\left(\; \boxed{-3,4} \; \right)\quad \textsf{(smaller $x$-value)}`

`(x,y)=\left(\; \boxed{5,0} \; \right)\quad \textsf{(larger $x$-value)}`

Explanation:

Given system of equations:

`\begin{cases}\;x+2y=5\\x^2+y^2=25\end{cases}`

To solve by the method of substitution, rearrange the first equation to make x the subject:

`\implies x=5-2y`

Substitute the found expression for x into the second equation and rearrange so that the equation equals zero:

`\begin{aligned}x=5-2y \implies (5-2y)^2+y^2&=25\\25-20y+4y^2+y^2&=25\\5y^2-20y+25&=25\\5y^2-20y&=0\end{aligned}`

Factor the equation:

`\begin{aligned}5y^2-20y&=0\\5y(y-4)&=0\end{aligned}`

Apply the zero-product property and solve for y:

`5y=0 \implies y=0`

`y-4=0 \implies y=4`

Substitute the found values of y into the first equation and solve for x:

`\begin{aligned}y=0 \implies x+2(0)&=5\\x&=5\end{aligned}`

`\begin{aligned}y=4 \implies x+2(4)&=5\\x+8&=5\\x&=-3\end{aligned}`

Therefore, the solutions are:

`(x,y)=\left(\; \boxed{-3,4} \; \right)\quad \textsf{(smaller $x$-value)}`

`(x,y)=\left(\; \boxed{5,0} \; \right)\quad \textsf{(larger $x$-value)}`

Answer 2

• (-3, 4)
• (5, 0)

=====================

Given system

• x + 2y = 5
• x² + y² = 25

Rearrange the first equation

• x = 5 - 2y

Substitute the value of x into second equation

• (5 - 2y)² + y² = 25
• 4y² - 20y + 25 + y² = 25
• 5y² - 20y = 0
• y² - 4y = 0
• y(y - 4) = 0
• y = 0 and y = 4

Find the value of x

• y = 0 ⇒ x = 5 - 2*0 = 5
• y = 4 ⇒ x = 5 - 2*4 = -3