Chapter 4: Q24. (page 157)

Solve each equation.
$[\begin{matrix} x^{2} + 1 & 5 - y \\ x + y & y - 4 \end{matrix}] = [\begin{matrix} 5 & x \\ 5 & 3 \end{matrix}]$

Short Answer

Expert verified

The solution is $(- 2, 7)$ .

Step by step solution

– Description of step.

As the matrices are equal, the corresponding elements are equal. Write the four linear equations to show this equality.

$\begin{matrix} x^{2} + 1 = 5 \\ 5 - y = x \\ x + y = 5 \\ y - 4 = 3 \end{matrix}$

– Description of step.

Add 4 on both the sides of $y - 4 = 3$ and simplify.

$\begin{matrix} y - 4 + 4 = 3 + 4 \\ y = 7 \end{matrix}$

– Description of step.

Substitute $7$ for $y$ into $5 - y = x$ and simplify.

$\begin{matrix} 5 - 7 = x \\ - 2 = x \end{matrix}$

Therefore, the solution is $(- 2, 7)$ .

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Solve each equation.
$[\begin{matrix} x^{2} + 1 & 5 - y \\ x + y & y - 4 \end{matrix}] = [\begin{matrix} 5 & x \\ 5 & 3 \end{matrix}]$

Short Answer

Step by step solution

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– Description of step.

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Solve each equation.x2+15-yx+yy-4=5x53

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Step by step solution

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Solve each equation.
$[\begin{matrix} x^{2} + 1 & 5 - y \\ x + y & y - 4 \end{matrix}] = [\begin{matrix} 5 & x \\ 5 & 3 \end{matrix}]$