NO LINKS!! Use the method of substitution to solve the system. (If there's no solution, enter no solution). Part 6z
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Answer:
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This gives us:
Answer:
[tex](x,y,z)=\left(\; \boxed{-1,7,-4} \; \right)\quad \textsf{(smaller $x$-value)}[/tex]
[tex](x,y,z)=\left(\; \boxed{1,-1,4} \; \right)\quad \textsf{(larger $x$-value)}[/tex]
Step-by-step explanation:
Given system of equations:
[tex]\begin{cases}x^2+z^2=17\\\;4x+y=3\\\;\;\;y+z=3\end{cases}[/tex]
To solve by the method of substitution, first rearrange the third equation to make y the subject:
[tex]\implies y=3-z[/tex]
Substitute this into the second equation and solve for z:
[tex]\begin{aligned}\implies 4x+(3-z)&=3\\3-z&=3-4x\\-z&=-4x\\z&=4x\end{aligned}[/tex]
Substitute the found expression for z into the first equation and solve for x:
[tex]\begin{aligned}\implies x^2+(4x)^2&=17\\x^2+16x^2&=17\\17x^2&=17\\x^2&=1\\x&=\pm1\end{aligned}[/tex]
Substitute the found values of x into the second equation and solve for y:
[tex]\begin{aligned}\implies x=-1 \implies 4(-1)+y&=3\\-4+y&=3\\y&=7\end{aligned}[/tex]
[tex]\begin{aligned}\implies x=1 \implies 4(1)+y&=3\\4+y&=3\\y&=-1\end{aligned}[/tex]
Substitute the found values of x into the derived expression for z and solve for z:
[tex]\begin{aligned}\implies x=-1 \implies z&=4(-1)\\z&=-4\end{aligned}[/tex]
[tex]\begin{aligned}\implies x=1 \implies z&=4(1)\\z&=4\end{aligned}[/tex]
Therefore, the solutions are:
[tex](x,y,z)=\left(\; \boxed{-1,7,-4} \; \right)\quad \textsf{(smaller $x$-value)}[/tex]
[tex](x,y,z)=\left(\; \boxed{1,-1,4} \; \right)\quad \textsf{(larger $x$-value)}[/tex]