Solve the System of Equations y=x^2+3x+4 y-x=7

$y = x^{2} + 3 x + 4$ $y - x = 7$

Replace all occurrences of $y$ with $x^{2} + 3 x + 4$ in each equation.

Replace all occurrences of $y$ in $y - x = 7$ with $x^{2} + 3 x + 4$ .

$(x^{2} + 3 x + 4) - x = 7$

$y = x^{2} + 3 x + 4$

Simplify the left side.

Subtract $x$ from $3 x$ .

$x^{2} + 2 x + 4 = 7$

$y = x^{2} + 3 x + 4$

$x^{2} + 2 x + 4 = 7$

$y = x^{2} + 3 x + 4$

$x^{2} + 2 x + 4 = 7$

$y = x^{2} + 3 x + 4$

Solve for $x$ in $x^{2} + 2 x + 4 = 7$ .

Subtract $7$ from both sides of the equation.

$x^{2} + 2 x + 4 - 7 = 0$

$y = x^{2} + 3 x + 4$

Subtract $7$ from $4$ .

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

$y = x^{2} + 3 x + 4$

Factor $x^{2} + 2 x - 3$ 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 $- 3$ and whose sum is $2$ .

$- 1, 3$

$y = x^{2} + 3 x + 4$

Write the factored form using these integers.

$(x - 1) (x + 3) = 0$

$y = x^{2} + 3 x + 4$

$(x - 1) (x + 3) = 0$

$y = x^{2} + 3 x + 4$

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

$x - 1 = 0$

$x + 3 = 0$

$y = x^{2} + 3 x + 4$

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

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

$x - 1 = 0$

$y = x^{2} + 3 x + 4$

Add $1$ to both sides of the equation.

$x = 1$

$y = x^{2} + 3 x + 4$

$x = 1$

$y = x^{2} + 3 x + 4$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

$y = x^{2} + 3 x + 4$

Subtract $3$ from both sides of the equation.

$x = - 3$

$y = x^{2} + 3 x + 4$

$x = - 3$

$y = x^{2} + 3 x + 4$

The final solution is all the values that make $(x - 1) (x + 3) = 0$ true.

$x = 1, - 3$

$y = x^{2} + 3 x + 4$

$x = 1, - 3$

$y = x^{2} + 3 x + 4$

Replace all occurrences of $x$ with $1$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} + 3 x + 4$ with $1$ .

$y = {(1)}^{2} + 3 (1) + 4$

$x = 1$

Simplify the right side.

Simplify ${(1)}^{2} + 3 (1) + 4$ .

Simplify each term.

One to any power is one.

$y = 1 + 3 (1) + 4$

$x = 1$

Multiply $3$ by $1$ .

$y = 1 + 3 + 4$

$x = 1$

$y = 1 + 3 + 4$

$x = 1$

Simplify by adding numbers.

Add $1$ and $3$ .

$y = 4 + 4$

$x = 1$

Add $4$ and $4$ .

$y = 8$

$x = 1$

$y = 8$

$x = 1$

$y = 8$

$x = 1$

$y = 8$

$x = 1$

$y = 8$

$x = 1$

Replace all occurrences of $x$ with $- 3$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} + 3 x + 4$ with $- 3$ .

$y = {(- 3)}^{2} + 3 (- 3) + 4$

$x = - 3$

Simplify the right side.

Simplify ${(- 3)}^{2} + 3 (- 3) + 4$ .

Simplify each term.

Raise $- 3$ to the power of $2$ .

$y = 9 + 3 (- 3) + 4$

$x = - 3$

Multiply $3$ by $- 3$ .

$y = 9 - 9 + 4$

$x = - 3$

$y = 9 - 9 + 4$

$x = - 3$

Simplify by adding and subtracting.

Subtract $9$ from $9$ .

$y = 0 + 4$

$x = - 3$

Add $0$ and $4$ .

$y = 4$

$x = - 3$

$y = 4$

$x = - 3$

$y = 4$

$x = - 3$

$y = 4$

$x = - 3$

$y = 4$

$x = - 3$

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

$(1, 8)$

$(- 3, 4)$

The result can be shown in multiple forms.

Point Form:

$(1, 8), (- 3, 4)$

Equation Form:

$x = 1, y = 8$

$x = - 3, y = 4$

Solve the System of Equations y=x^2+3x+4 y-x=7

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