Solve the System of Equations x^2-y=7 y-5x=-11

$x^{2} - y = 7$ $y - 5 x = - 11$

Step 1

Add $5 x$ to both sides of the equation.

$y = - 11 + 5 x$

$x^{2} - y = 7$

Step 2

Replace all occurrences of $y$ with $- 11 + 5 x$ in each equation.

Replace all occurrences of $y$ in $x^{2} - y = 7$ with $- 11 + 5 x$ .

$x^{2} - (- 11 + 5 x) = 7$

$y = - 11 + 5 x$

Simplify the left side.

Simplify each term.

Apply the distributive property.

$x^{2} + 11 - (5 x) = 7$

$y = - 11 + 5 x$

Multiply $- 1$ by $- 11$ .

$x^{2} + 11 - (5 x) = 7$

$y = - 11 + 5 x$

Multiply $5$ by $- 1$ .

$x^{2} + 11 - 5 x = 7$

$y = - 11 + 5 x$

$x^{2} + 11 - 5 x = 7$

$y = - 11 + 5 x$

$x^{2} + 11 - 5 x = 7$

$y = - 11 + 5 x$

$x^{2} + 11 - 5 x = 7$

$y = - 11 + 5 x$

Step 3

Solve for $x$ in $x^{2} + 11 - 5 x = 7$ .

Subtract $7$ from both sides of the equation.

$x^{2} + 11 - 5 x - 7 = 0$

$y = - 11 + 5 x$

Subtract $7$ from $11$ .

$x^{2} - 5 x + 4 = 0$

$y = - 11 + 5 x$

Factor $x^{2} - 5 x + 4$ 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 $4$ and whose sum is $- 5$ .

$- 4, - 1$

$y = - 11 + 5 x$

Write the factored form using these integers.

$(x - 4) (x - 1) = 0$

$y = - 11 + 5 x$

$(x - 4) (x - 1) = 0$

$y = - 11 + 5 x$

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

$x - 4 = 0$

$x - 1 = 0$

$y = - 11 + 5 x$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

$y = - 11 + 5 x$

Add $4$ to both sides of the equation.

$x = 4$

$y = - 11 + 5 x$

$x = 4$

$y = - 11 + 5 x$

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

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

$x - 1 = 0$

$y = - 11 + 5 x$

Add $1$ to both sides of the equation.

$x = 1$

$y = - 11 + 5 x$

$x = 1$

$y = - 11 + 5 x$

The final solution is all the values that make $(x - 4) (x - 1) = 0$ true.

$x = 4, 1$

$y = - 11 + 5 x$

$x = 4, 1$

$y = - 11 + 5 x$

Step 4

Replace all occurrences of $x$ with $4$ in each equation.

Replace all occurrences of $x$ in $y = - 11 + 5 x$ with $4$ .

$y = - 11 + 5 (4)$

$x = 4$

Simplify the right side.

Simplify $- 11 + 5 (4)$ .

Multiply $5$ by $4$ .

$y = - 11 + 20$

$x = 4$

Add $- 11$ and $20$ .

$y = 9$

$x = 4$

$y = 9$

$x = 4$

$y = 9$

$x = 4$

$y = 9$

$x = 4$

Step 5

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

Replace all occurrences of $x$ in $y = - 11 + 5 x$ with $1$ .

$y = - 11 + 5 (1)$

$x = 1$

Simplify the right side.

Simplify $- 11 + 5 (1)$ .

Multiply $5$ by $1$ .

$y = - 11 + 5$

$x = 1$

Add $- 11$ and $5$ .

$y = - 6$

$x = 1$

$y = - 6$

$x = 1$

$y = - 6$

$x = 1$

$y = - 6$

$x = 1$

Step 6

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

$(4, 9)$

$(1, - 6)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(4, 9), (1, - 6)$

Equation Form:

$x = 4, y = 9$

$x = 1, y = - 6$

Step 8

Solve the System of Equations x^2-y=7 y-5x=-11

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