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University of California - Los Angeles

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Problem 9

In Exercises 7-10, find $x$ and $y$.

$\left[ \begin{array}{r} 16 & 4 & 5 & 4 \\ -3 & 13 & 15 & 6 \\ 0 & 2 & 4 & 0 \end{array} \right] = \left[ \begin{array}{r} 16 & 4 & 2x+1 & 4 \\ -3 & 13 & 15 & 3x \\ 0 & 2 & 3y-5 & 0 \end{array} \right]$

Answer

$x=2, y=3$

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## Discussion

## Video Transcript

The question here gives us two different matrices, as seen by the 1st 1 over here on the 2nd 1 over here. And it basically states that as the correspondent entries in each of these aren't the same as we can see by the first row in the first calm of each, Um, we can say that these two major cities are there for equal to each other. So from here, we want us. It wants us tow. Find the values of accent. Why So, first of all, if we look at the third row in the first column over here, we can initially solve this to find, um, the value of X. So if we set these equal to each other, five is equal to two backs plus one we can solve it for X is equal to two. As if we move this over here that before and we divide everything. My two. That would be too. Um, this could be rechecked by looking at the, um, entry over at the fourth row second column Over here, three x is equal to, I mean to six, rather, which basically also states that X is equal to two for the values of why, If we look at the third row third column over here on both Mitrice's weaken, set these equal to each other. Three y minus five. Busy quits of four entrances. Sulfur wise three y is equal to nine. Therefore why? As he puts a three. So from here, we know that why is equal to three X is equal to two in this particular matrix.

## Recommended Questions

In Exercises 7-10, find $x$ and $y$.

$\left[ \begin{array}{r} x+2 & 8 & -3 \\ 1 & 2y & 2x \\ 7 & -2 & y+2 \end{array} \right] = \left[ \begin{array}{r} 2x+6 & 8 & -3 \\ 1 & 18 & -8 \\ 7 & -2 & 11 \end{array} \right]$

In Exercises 7-10, find $x$ and $y$.

$\left[ \begin{array}{r} x & -2 \\ 7 & y \end{array} \right] = \left[ \begin{array}{r} -4 & -2 \\ 7 & 22 \end{array} \right]$

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$$

\left[\begin{array}{cccc}{16} & {4} & {5} & {4} \\ {-3} & {13} & {15} & {12} \\ {0} & {2} & {4} & {0}\end{array}\right]=\left[\begin{array}{cccc}{16} & {4} & {2 x+7} & {4} \\ {-3} & {13} & {15} & {3 y} \\ {0} & {2} & {3 z-14} & {0}\end{array}\right]

$$

Solve each equation in Exercises 41β60 by making an appropriate substitution.

$$\left(y-\frac{10}{y}\right)^{2}+6\left(y-\frac{10}{y}\right)-27=0$$

In Exercises $1-10$, determine whether the given ordered pair is a solution of the system.

$$\begin{aligned}

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&\left\{\begin{array}{l}

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In Exercises 1β4, find x and y or x, y, and z.

$$

\left[\begin{array}{rrr}{x+4} & {8} & {-3} \\ {1} & {22} & {2 y} \\ {7} & {-2} & {z+2}\end{array}\right]=\left[\begin{array}{rrr}{2 x+9} & {8} & {-3} \\ {1} & {22} & {-8} \\ {7} & {-2} & {11}\end{array}\right]

$$

In Exercises 111β113, perform the indicated operations.

$$

[(7 x+5)+4 y][(7 x+5)-4 y]

$$

In the following exercises, solve each equation.

$$

3(y-5)-2 y=-7

$$

In Exercises $15-82,$ find each product.

$$

(3 x+7-5 y)(3 x+7+5 y)

$$