Solve the System of Equations y=6x+3 y=x^2+x-3

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

Eliminate the equal sides of each equation and combine.

$6 x + 3 = x^{2} + x - 3$

Solve $6 x + 3 = x^{2} + x - 3$ for $x$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$x^{2} + x - 3 = 6 x + 3$

Move all terms containing $x$ to the left side of the equation.

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

$x^{2} + x - 3 - 6 x = 3$

Subtract $6 x$ from $x$ .

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

Subtract $3$ from both sides of the equation.

$x^{2} - 5 x - 3 - 3 = 0$

Subtract $3$ from $- 3$ .

$x^{2} - 5 x - 6 = 0$

Factor $x^{2} - 5 x - 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 6$ and whose sum is $- 5$ .

$- 6, 1 + y = x^{2} + x - 3$

Write the factored form using these integers.

$(x - 6) (x + 1) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 6 = 0$

$x + 1 = 0 + y = x^{2} + x - 3$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

The final solution is all the values that make $(x - 6) (x + 1) = 0$ true.

$x = 6, - 1$

Evaluate $y$ when $x = 6$ .

Substitute $6$ for $x$ .

$y = {(6)}^{2} + 6 - 3$

Substitute $6$ for $x$ in $y = {(6)}^{2} + 6 - 3$ and solve for $y$ .

Remove parentheses.

$y = {(6)}^{2} + 6 - 3$

Remove parentheses.

$y = 6^{2} + 6 - 3$

Remove parentheses.

$y = {(6)}^{2} + 6 - 3$

Simplify ${(6)}^{2} + 6 - 3$ .

Raise $6$ to the power of $2$ .

$y = 36 + 6 - 3$

Add $36$ and $6$ .

$y = 42 - 3$

Subtract $3$ from $42$ .

$y = 39$

Evaluate $y$ when $x = - 1$ .

Substitute $- 1$ for $x$ .

$y = {(- 1)}^{2} - 1 - 3$

Substitute $- 1$ for $x$ in $y = {(- 1)}^{2} - 1 - 3$ and solve for $y$ .

Remove parentheses.

$y = {(- 1)}^{2} - 1 - 3$

Simplify ${(- 1)}^{2} - 1 - 3$ .

Raise $- 1$ to the power of $2$ .

$y = 1 - 1 - 3$

Subtract $1$ from $1$ .

$y = 0 - 3$

Subtract $3$ from $0$ .

$y = - 3$

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

$(6, 39)$

$(- 1, - 3)$

The result can be shown in multiple forms.

Point Form:

$(6, 39), (- 1, - 3)$

Equation Form:

$x = 6, y = 39$

$x = - 1, y = - 3$

Solve the System of Equations y=6x+3 y=x^2+x-3

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