The answer is:
Solve for y in the first equation
y=−5<span>
y=8−<span>x
</span></span>
<span>Replace all occurrences of y with the solution found by solving the last equation for y. In this case, the value substituted is <span>−5</span></span><span>y=−5</span><span><span>(−5)</span>=8−<span>x
</span></span>
Remove parentheses.<span>y=−5</span><span>−5=8−<span>x
</span></span>
Solve for x in the second equation.<span>y=−5</span>
<span>x=13</span>
The answer is <span>(13,−5<span>) Please mark as Brainliest
</span></span>
Answer:
x=4, y=4, λ=-16
Step-by-step explanation:
We have this 3x3 system of linear equations:
λ
λ

So, let's rewrite the system in its augmented matrix form
![\left[\begin{array}{cccc}4&0&1&0\\0&4&1&0\\1&1&0&8\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D4%260%261%260%5C%5C0%264%261%260%5C%5C1%261%260%268%5Cend%7Barray%7D%5Cright%5D)
Let´s apply row reduction process to its associated augmented matrix:
Swap R1 and R3
![\left[\begin{array}{cccc}1&1&0&8\\0&4&1&0\\4&0&1&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%260%268%5C%5C0%264%261%260%5C%5C4%260%261%260%5Cend%7Barray%7D%5Cright%5D)
R3-4R1
![\left[\begin{array}{cccc}1&1&0&8\\0&4&1&0\\0&-4&1&-32\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%260%268%5C%5C0%264%261%260%5C%5C0%26-4%261%26-32%5Cend%7Barray%7D%5Cright%5D)
R3+R2
![\left[\begin{array}{cccc}1&1&0&8\\0&4&1&0\\0&0&2&-32\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%261%260%268%5C%5C0%264%261%260%5C%5C0%260%262%26-32%5Cend%7Barray%7D%5Cright%5D)
Now we have a simplified system:
x+y+0=0
0+4y+λ=0
0+0+2λ=-32
Solving for λ, x, and y
λ=-16
x=4
y=4
Answer:
2.133333
Step-by-step explanation:
He didn't simplify it
Answer:
there positive
Step-by-step explanation:
Answer:
it is x>4 I solved it out.