Precalculus Examples

$12 x + 18 y = 5$ , $4 x - 3 y = 2$ , $(\frac{3}{4}, \frac{1}{3})$

Step 1

Simplify $12 (\frac{3}{4}) + 18 (\frac{1}{3})$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

Cancel the common factor of $4$ .

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Step 1.1.1.1

Factor $4$ out of $12$ .

$4 (3) (\frac{3}{4}) + 18 (\frac{1}{3}) = 5$

$4 (\frac{3}{4}) - 3 (\frac{1}{3}) = 2$

Step 1.1.1.2

Cancel the common factor.

$4 \cdot (3 (\frac{3}{4})) + 18 (\frac{1}{3}) = 5$

$4 (\frac{3}{4}) - 3 (\frac{1}{3}) = 2$

Step 1.1.1.3

Rewrite the expression.

$3 \cdot 3 + 18 (\frac{1}{3}) = 5$

$4 (\frac{3}{4}) - 3 (\frac{1}{3}) = 2$

$3 \cdot 3 + 18 (\frac{1}{3}) = 5$

$4 (\frac{3}{4}) - 3 (\frac{1}{3}) = 2$

Step 1.1.2

Multiply $3$ by $3$ .

$9 + 18 (\frac{1}{3}) = 5$

$4 (\frac{3}{4}) - 3 (\frac{1}{3}) = 2$

Step 1.1.3

Cancel the common factor of $3$ .

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Step 1.1.3.1

Factor $3$ out of $18$ .

$9 + 3 (6) (\frac{1}{3}) = 5$

$4 (\frac{3}{4}) - 3 (\frac{1}{3}) = 2$

Step 1.1.3.2

Cancel the common factor.

$9 + 3 \cdot (6 (\frac{1}{3})) = 5$

$4 (\frac{3}{4}) - 3 (\frac{1}{3}) = 2$

Step 1.1.3.3

Rewrite the expression.

$9 + 6 = 5$

$4 (\frac{3}{4}) - 3 (\frac{1}{3}) = 2$

$9 + 6 = 5$

$4 (\frac{3}{4}) - 3 (\frac{1}{3}) = 2$

$9 + 6 = 5$

$4 (\frac{3}{4}) - 3 (\frac{1}{3}) = 2$

Step 1.2

Add $9$ and $6$ .

$15 = 5$

$4 (\frac{3}{4}) - 3 (\frac{1}{3}) = 2$

$15 = 5$

$4 (\frac{3}{4}) - 3 (\frac{1}{3}) = 2$

Step 2

Since $15 \neq 5$ , there are no solutions.

No solution

Step 3

Simplify $4 (\frac{3}{4}) - 3 (\frac{1}{3})$ .

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Step 3.1

Simplify each term.

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Step 3.1.1

Cancel the common factor of $4$ .

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Step 3.1.1.1

Cancel the common factor.

$4 (\frac{3}{4}) - 3 (\frac{1}{3}) = 2$

No solution

Step 3.1.1.2

Rewrite the expression.

$3 - 3 (\frac{1}{3}) = 2$

No solution

$3 - 3 (\frac{1}{3}) = 2$

No solution

Step 3.1.2

Cancel the common factor of $3$ .

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Step 3.1.2.1

Factor $3$ out of $- 3$ .

$3 + 3 (- 1) (\frac{1}{3}) = 2$

No solution

Step 3.1.2.2

Cancel the common factor.

$3 + 3 \cdot (- 1 (\frac{1}{3})) = 2$

No solution

Step 3.1.2.3

Rewrite the expression.

$3 - 1 = 2$

No solution

$3 - 1 = 2$

No solution

$3 - 1 = 2$

No solution

Step 3.2

Subtract $1$ from $3$ .

$2 = 2$

No solution

No solution

Step 4

Since $2 = 2$ , the equation will always be true.

No solution

Step 5

The ordered pair is not a solution to the system of equations.

$(\frac{3}{4}, \frac{1}{3})$ is not a solution

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$