Answer: a = -5/2
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Work Shown:
The scalar (2/5a+6) multiplies with the matrix on the left side to get the matrix on the right side
So this is what your steps could look like
[tex](\frac{2}{5}a+6)\begin{bmatrix}3 \\ -2 \\ 7\end{bmatrix}=\begin{bmatrix}15 \\ -10 \\ 35\end{bmatrix}\\\\\\\begin{bmatrix}(\frac{2}{5}a+6)*3 \\ (\frac{2}{5}a+6)*(-2) \\ (\frac{2}{5}a+6)*7\end{bmatrix}=\begin{bmatrix}15 \\ -10 \\ 35\end{bmatrix}\\\\\\[/tex]
[tex]\begin{bmatrix}\frac{2}{5}a*3+6*3 \\ \frac{2}{5}a*(-2)+6*(-2) \\ \frac{2}{5}a*7+6*7\end{bmatrix}=\begin{bmatrix}15 \\ -10 \\ 35\end{bmatrix}\\\\\\\begin{bmatrix}\frac{6}{5}a+18 \\ -\frac{4}{5}a-12 \\ \frac{14}{5}a+42\end{bmatrix}=\begin{bmatrix}15 \\ -10 \\ 35\end{bmatrix}\\\\\\[/tex]
The last matrix equation leads to this system of equations
[tex]\begin{cases}\frac{6}{5}a+18=15 \\ -\frac{4}{5}a-12=-10 \\ \frac{14}{5}a+42=35\end{cases}\\\\\\[/tex]
If we pick any of those equations, and solve for 'a', then we'll get our answer.
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Let's say we pick the first equation
(6/5)a+18 = 15
(6/5)a = 15-18
(6/5)a = -3
a = -3(5/6)
a = -15/6
a = -5/2
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If we pick on the second equation, then,
(-4/5)a-12 = -10
(-4/5)a = -10+12
(-4/5)a = 2
a = 2(-5/4)
a = -10/4
a = -5/2
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Or we could solve the third equation for 'a'
(14/5)a+42 = 35
(14/5)a = 35-42
(14/5)a = -7
a = -7*(5/14)
a = -35/14
a = -5/2
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You only need to solve one equation to find 'a', though it's good practice to solve all three to see all three rows agreeing with one another.
