Find the Equation Using Slope-Intercept Form (1,7) 3x+7y=1

$(1, 7)$ $3 x + 7 y = 1$

Step 1

Solve for $m$ .

Subtract $7 y$ from both sides of the equation.

$3 x = 1 - 7 y$

Divide each term in $3 x = 1 - 7 y$ by $3$ and simplify.

Divide each term in $3 x = 1 - 7 y$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3} + \frac{- 7 y}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3} + \frac{- 7 y}{3}$

Divide $x$ by $1$ .

$x = \frac{1}{3} + \frac{- 7 y}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = \frac{1}{3} - \frac{7 y}{3}$

Step 2

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

Use the formula for the equation of a line to find $b$ .

$y = m x + b$

Substitute the value of $m$ into the equation.

$y = (\frac{1}{3} - \frac{7 y}{3}) \cdot x + b$

Substitute the value of $x$ into the equation.

$y = (\frac{1}{3} - \frac{7 y}{3}) \cdot (1) + b$

Substitute the value of $y$ into the equation.

$7 = (\frac{1}{3} - \frac{7 (7)}{3}) \cdot (1) + b$

Find the value of $b$ .

Rewrite the equation as $(\frac{1}{3} - \frac{7 (7)}{3}) \cdot 1 + b = 7$ .

$(\frac{1}{3} - \frac{7 (7)}{3}) \cdot 1 + b = 7$

Simplify each term.

Combine the numerators over the common denominator.

$\frac{1 - 7 \cdot 7}{3} \cdot 1 + b = 7$

Multiply $- 7$ by $7$ .

$\frac{1 - 49}{3} \cdot 1 + b = 7$

Subtract $49$ from $1$ .

$\frac{- 48}{3} \cdot 1 + b = 7$

Divide $- 48$ by $3$ .

$- 16 \cdot 1 + b = 7$

Multiply $- 16$ by $1$ .

$- 16 + b = 7$

Move all terms not containing $b$ to the right side of the equation.

Add $16$ to both sides of the equation.

$b = 7 + 16$

Add $7$ and $16$ .

$b = 23$

Step 3

Now that the values of $m$ (slope) and $b$ (y-intercept) are known, substitute them into $y = m x + b$ to find the equation of the line.

$y = \frac{1}{3} x - \frac{7}{3} x y + 23$

Step 4

Find the Equation Using Slope-Intercept Form (1,7) 3x+7y=1

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