Solve by Graphing y=x^2-5x+3 6x-y=27

$y = x^{2} - 5 x + 3$ $6 x - y = 27$

Step 1

Subtract $6 x$ from both sides of the equation.

$- y = 27 - 6 x$

$y = x^{2} - 5 x + 3$

Step 2

Divide each term in $- y = 27 - 6 x$ by $- 1$ and simplify.

Divide each term in $- y = 27 - 6 x$ by $- 1$ .

$\frac{- y}{- 1} = \frac{27}{- 1} + \frac{- 6 x}{- 1}$

$y = x^{2} - 5 x + 3$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{27}{- 1} + \frac{- 6 x}{- 1}$

$y = x^{2} - 5 x + 3$

Divide $y$ by $1$ .

$y = \frac{27}{- 1} + \frac{- 6 x}{- 1}$

$y = x^{2} - 5 x + 3$

$y = \frac{27}{- 1} + \frac{- 6 x}{- 1}$

$y = x^{2} - 5 x + 3$

Simplify the right side.

Simplify each term.

Divide $27$ by $- 1$ .

$y = - 27 + \frac{- 6 x}{- 1}$

$y = x^{2} - 5 x + 3$

Move the negative one from the denominator of $\frac{- 6 x}{- 1}$ .

$y = - 27 - 1 \cdot (- 6 x)$

$y = x^{2} - 5 x + 3$

Rewrite $- 1 \cdot (- 6 x)$ as $- (- 6 x)$ .

$y = - 27 - (- 6 x)$

$y = x^{2} - 5 x + 3$

Multiply $- 6$ by $- 1$ .

$y = - 27 + 6 x$

$y = x^{2} - 5 x + 3$

$y = - 27 + 6 x$

$y = x^{2} - 5 x + 3$

$y = - 27 + 6 x$

$y = x^{2} - 5 x + 3$

$y = - 27 + 6 x$

$y = x^{2} - 5 x + 3$

Step 3

Create a graph to locate the intersection of the equations. The intersection of the system of equations is the solution.

$(6, 9)$

$(5, 3)$

Step 4