Algebra Examples

What is the equation of the line that passes through the point

(- 5, 2)

$(- 5, 2)$ and has a slope of

\frac{4}{5}

$\frac{4}{5}$ ?

Step 1

Find the value of

b

$b$ using the formula for the equation of a line.

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Step 1.1

Use the formula for the equation of a line to find

b

$b$ .

y = m x + b

$y = m x + b$

Step 1.2

Substitute the value of

m

$m$ into the equation.

y = (\frac{4}{5}) x + b

$y = (\frac{4}{5}) x + b$

Step 1.3

Substitute the value of

x

$x$ into the equation.

y = (\frac{4}{5}) \cdot (- 5) + b

$y = (\frac{4}{5}) \cdot (- 5) + b$

Step 1.4

Substitute the value of

y

$y$ into the equation.

2 = (\frac{4}{5}) \cdot (- 5) + b

$2 = (\frac{4}{5}) \cdot (- 5) + b$

Step 1.5

Find the value of

b

$b$ .

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Step 1.5.1

Rewrite the equation as

\frac{4}{5} \cdot - 5 + b = 2

$\frac{4}{5} \cdot - 5 + b = 2$ .

\frac{4}{5} \cdot - 5 + b = 2

$\frac{4}{5} \cdot - 5 + b = 2$

Step 1.5.2

Simplify each term.

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Step 1.5.2.1

Cancel the common factor of

5

$5$ .

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Step 1.5.2.1.1

Factor

5

$5$ out of

- 5

$- 5$ .

\frac{4}{5} \cdot (5 (- 1)) + b = 2

$\frac{4}{5} \cdot (5 (- 1)) + b = 2$

Step 1.5.2.1.2

Cancel the common factor.

\frac{4}{5} \cdot (5 \cdot - 1) + b = 2

$\frac{4}{5} \cdot (5 \cdot - 1) + b = 2$

Step 1.5.2.1.3

Rewrite the expression.

4 \cdot - 1 + b = 2

$4 \cdot - 1 + b = 2$

4 \cdot - 1 + b = 2

$4 \cdot - 1 + b = 2$

Step 1.5.2.2

Multiply

4

$4$ by

- 1

$- 1$ .

- 4 + b = 2

$- 4 + b = 2$

- 4 + b = 2

$- 4 + b = 2$

Step 1.5.3

Move all terms not containing

b

$b$ to the right side of the equation.

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Step 1.5.3.1

Add

4

$4$ to both sides of the equation.

b = 2 + 4

$b = 2 + 4$

Step 1.5.3.2

Add

2

$2$ and

4

$4$ .

b = 6

$b = 6$

b = 6

$b = 6$

b = 6

$b = 6$

b = 6

$b = 6$

Step 2

Now that the values of

m

$m$ (slope) and

b

$b$ (y-intercept) are known, substitute them into

y = m x + b

$y = m x + b$ to find the equation of the line.

y = \frac{4}{5} x + 6

$y = \frac{4}{5} x + 6$

Step 3

⎡ ⎢ ⎣ \begin{matrix} x^{2} & \frac{1}{2} \sqrt{π} & \int x d x \end{matrix} ⎤ ⎥ ⎦

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