Trigonometry Examples

$y = 2 x - 3$ , $y = - \frac{1}{3} x + 4$ , $x = 9$

Step 1

Replace all occurrences of $x$ with $9$ in each equation.

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Step 1.1

Replace all occurrences of $x$ in $y = 2 x - 3$ with $9$ .

$y = 2 (9) - 3$

$x = 9$

$y = - \frac{1}{3} x + 4$

Step 1.2

Simplify the right side.

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Step 1.2.1

Simplify $2 (9) - 3$ .

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Step 1.2.1.1

Multiply $2$ by $9$ .

$y = 18 - 3$

$x = 9$

$y = - \frac{1}{3} x + 4$

Step 1.2.1.2

Subtract $3$ from $18$ .

$y = 15$

$x = 9$

$y = - \frac{1}{3} x + 4$

$y = 15$

$x = 9$

$y = - \frac{1}{3} x + 4$

$y = 15$

$x = 9$

$y = - \frac{1}{3} x + 4$

Step 1.3

Replace all occurrences of $x$ in $y = - \frac{1}{3} x + 4$ with $9$ .

$y = - \frac{1}{3} \cdot 9 + 4$

$y = 15$

$x = 9$

Step 1.4

Simplify the right side.

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Step 1.4.1

Simplify $- \frac{1}{3} \cdot 9 + 4$ .

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Step 1.4.1.1

Simplify each term.

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Step 1.4.1.1.1

Cancel the common factor of $3$ .

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Step 1.4.1.1.1.1

Move the leading negative in $- \frac{1}{3}$ into the numerator.

$y = \frac{- 1}{3} \cdot 9 + 4$

$y = 15$

$x = 9$

Step 1.4.1.1.1.2

Factor $3$ out of $9$ .

$y = \frac{- 1}{3} \cdot (3 (3)) + 4$

$y = 15$

$x = 9$

Step 1.4.1.1.1.3

Cancel the common factor.

$y = \frac{- 1}{3} \cdot (3 \cdot 3) + 4$

$y = 15$

$x = 9$

Step 1.4.1.1.1.4

Rewrite the expression.

$y = - 1 \cdot 3 + 4$

$y = 15$

$x = 9$

$y = - 1 \cdot 3 + 4$

$y = 15$

$x = 9$

Step 1.4.1.1.2

Multiply $- 1$ by $3$ .

$y = - 3 + 4$

$y = 15$

$x = 9$

$y = - 3 + 4$

$y = 15$

$x = 9$

Step 1.4.1.2

Add $- 3$ and $4$ .

$y = 1$

$y = 15$

$x = 9$

$y = 1$

$y = 15$

$x = 9$

$y = 1$

$y = 15$

$x = 9$

$y = 1$

$y = 15$

$x = 9$

Step 2

The solution to the system is the complete set of ordered pairs that are valid solutions.

$(9, 15)$

Step 3

The result can be shown in multiple forms.

Point Form:

$(9, 15)$

Equation Form:

$x = 9, y = 15$

Step 4