Precalculus Examples

$x^{2} + 3 y^{2} + 4 y = 14$ , $(5, - 2)$

Step 1

Insert the values of $x = 5$ and $y = - 2$ into the equation to find if the ordered pair is a solution.

${(5)}^{2} + 3 {(- 2)}^{2} + 4 (- 2) = 14$

Step 2

Simplify $5^{2} + 3 {(- 2)}^{2} + 4 (- 2)$ .

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Step 2.1

Simplify each term.

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Step 2.1.1

Raise $5$ to the power of $2$ .

$25 + 3 {(- 2)}^{2} + 4 (- 2) = 14$

Step 2.1.2

Raise $- 2$ to the power of $2$ .

$25 + 3 \cdot 4 + 4 (- 2) = 14$

Step 2.1.3

Multiply $3$ by $4$ .

$25 + 12 + 4 (- 2) = 14$

Step 2.1.4

Multiply $4$ by $- 2$ .

$25 + 12 - 8 = 14$

Step 2.2

Simplify by adding and subtracting.

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Step 2.2.1

Add $25$ and $12$ .

$37 - 8 = 14$

Step 2.2.2

Subtract $8$ from $37$ .

$29 = 14$

Step 3

Since $29 \neq 14$ , there are no solutions.

No solution

Step 4

Since the equation is not true when the values are used, the ordered pair is not a solution.

The ordered pair is not a solution to the equation.