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4 years ago

What is the value of k? 2k+9=7−3k

2 answers:

Elena-2011 [213]4 years ago

6 0

2k + 9 = 7 - 3k

2k + 9 - 7 = -3k

2k + 2 = -3k

2 = -3k - 2k

2 = -5k

-2/5 = k

k = -2/5

hope that helps, God bless!

Andrews [41]4 years ago

3 0

-2/5

sssssssssssssssssssssss

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4 years ago

Jacob solves the system of equations by forming a matrix equation.

Lyrx [107]

Simultaneous equations can be solved using inverse matrix operation.

The complete steps of Jacob's solution are:

$\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]^{-1} \cdot \left[\begin{array}{cc}4&1\\-2&3\end{array}\right]\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14}\left[\begin{array}{cc}3&-1\\2&4\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}2&-6\end{array}\right]$

We have:

$4x + y = 2$

$-2x + 4y = -22$

Calculate the determinant of $\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]$

$|A| = 4 \times 3 -1 \times -2$

$|A| = 12 +2$

$|A| = 14$

So, the inverse matrix becomes

$A = \frac{1}{14}\left[\begin{array}{cc}4&1\\-2&3\end{array}\right]$

Replace the first column with $\left[\begin{array}{c}2&-22\end{array}\right]$ to calculate the value of x

$x = \frac{1}{14}\left[\begin{array}{cc}2&1\\-22&3\end{array}\right]$

So, we have:

$x = \frac{1}{14}(2 \times 3 - 1 \times -22)$

$x = \frac{1}{14}(6 +22)$

$x = \frac{1}{14}(28)$

$x = 2$

Replace the second column with $\left[\begin{array}{c}2&-22\end{array}\right]$ to calculate the value of y

$y = \frac{1}{14}\left[\begin{array}{cc}4&2\\-2&-22\end{array}\right]$

So, we have:

$y = \frac{1}{14}(4 \times -22 - 2 \times -2)$

$y = \frac{1}{14}(-88 +4)$

$y = \frac{1}{14}(-84)$

$y = -6$

Hence, the complete process is:

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{cc}4&1\\-2&3\end{array}\right] \cdot \left[\begin{array}{c}2&-22\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \frac{1}{14} \left[\begin{array}{c}28&-84\end{array}\right]$

$\left[\begin{array}{c}x&y\end{array}\right] = \left[\begin{array}{c}2&-6\end{array}\right]$