Question

NO LINKS!! Use the method of substitution to solve the system. (If there's no solution, enter no solution). Part 6z​

Answer 1

`(x,y,z)=\left(\; \boxed{-1,7,-4} \; \right)\quad \textsf{(smaller $x$-value)}`

`(x,y,z)=\left(\; \boxed{1,-1,4} \; \right)\quad \textsf{(larger $x$-value)}`

Explanation:

Given system of equations:

`\begin{cases}x^2+z^2=17\\\;4x+y=3\\\;\;\;y+z=3\end{cases}`

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

`\implies y=3-z`

Substitute this into the second equation and solve for z:

`\begin{aligned}\implies 4x+(3-z)&=3\\3-z&=3-4x\\-z&=-4x\\z&=4x\end{aligned}`

Substitute the found expression for z into the first equation and solve for x:

`\begin{aligned}\implies x^2+(4x)^2&=17\\x^2+16x^2&=17\\17x^2&=17\\x^2&=1\\x&=\pm1\end{aligned}`

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

`\begin{aligned}\implies x=-1 \implies 4(-1)+y&=3\\-4+y&=3\\y&=7\end{aligned}`

`\begin{aligned}\implies x=1 \implies 4(1)+y&=3\\4+y&=3\\y&=-1\end{aligned}`

Substitute the found values of x into the derived expression for z and solve for z:

`\begin{aligned}\implies x=-1 \implies z&=4(-1)\\z&=-4\end{aligned}`

`\begin{aligned}\implies x=1 \implies z&=4(1)\\z&=4\end{aligned}`

Therefore, the solutions are:

`(x,y,z)=\left(\; \boxed{-1,7,-4} \; \right)\quad \textsf{(smaller $x$-value)}`

`(x,y,z)=\left(\; \boxed{1,-1,4} \; \right)\quad \textsf{(larger $x$-value)}`

Answer 2

• (-1, 7, - 4)
• (1, -1, 4)

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

Given system

• x² + z² = 17
• 4x + y = 3
• y + z = 3

Rearrange the last two equation

• 4x = 3 - y
• z = 3 - y

This gives us:

• z = 4x

Substitute the value of z into fist equation

• x² + (4x)² = 17
• x² + 16x² = 17
• 17x² = 17
• x² = 1
• x = 1 and x = - 1

Find values of z and y

• x = 1 ⇒ z = 4*1 = 4 ⇒ y = 3 - 4 = - 1
• x = - 1 ⇒ z = 4*(-1) = - 4 ⇒ y = 3 - (-4) = 7