Solve the System of Equations y=x^2-10x+12 2x+y=5

$y = x^{2} - 10 x + 12$ $2 x + y = 5$

Step 1

Replace all occurrences of $y$ with $x^{2} - 10 x + 12$ in each equation.

Replace all occurrences of $y$ in $2 x + y = 5$ with $x^{2} - 10 x + 12$ .

$2 x + x^{2} - 10 x + 12 = 5$

$y = x^{2} - 10 x + 12$

Simplify the left side.

Simplify $2 x + x^{2} - 10 x + 12$ .

Remove parentheses.

$2 x + x^{2} - 10 x + 12 = 5$

$y = x^{2} - 10 x + 12$

Subtract $10 x$ from $2 x$ .

$x^{2} - 8 x + 12 = 5$

$y = x^{2} - 10 x + 12$

$x^{2} - 8 x + 12 = 5$

$y = x^{2} - 10 x + 12$

$x^{2} - 8 x + 12 = 5$

$y = x^{2} - 10 x + 12$

$x^{2} - 8 x + 12 = 5$

$y = x^{2} - 10 x + 12$

Step 2

Solve for $x$ in $x^{2} - 8 x + 12 = 5$ .

Subtract $5$ from both sides of the equation.

$x^{2} - 8 x + 12 - 5 = 0$

$y = x^{2} - 10 x + 12$

Subtract $5$ from $12$ .

$x^{2} - 8 x + 7 = 0$

$y = x^{2} - 10 x + 12$

Factor $x^{2} - 8 x + 7$ 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 $7$ and whose sum is $- 8$ .

$- 7, - 1$

$y = x^{2} - 10 x + 12$

Write the factored form using these integers.

$(x - 7) (x - 1) = 0$

$y = x^{2} - 10 x + 12$

$(x - 7) (x - 1) = 0$

$y = x^{2} - 10 x + 12$

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

$x - 7 = 0$

$x - 1 = 0$

$y = x^{2} - 10 x + 12$

Set $x - 7$ equal to $0$ and solve for $x$ .

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

$y = x^{2} - 10 x + 12$

Add $7$ to both sides of the equation.

$x = 7$

$y = x^{2} - 10 x + 12$

$x = 7$

$y = x^{2} - 10 x + 12$

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

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

$x - 1 = 0$

$y = x^{2} - 10 x + 12$

Add $1$ to both sides of the equation.

$x = 1$

$y = x^{2} - 10 x + 12$

$x = 1$

$y = x^{2} - 10 x + 12$

The final solution is all the values that make $(x - 7) (x - 1) = 0$ true.

$x = 7, 1$

$y = x^{2} - 10 x + 12$

$x = 7, 1$

$y = x^{2} - 10 x + 12$

Step 3

Replace all occurrences of $x$ with $7$ in each equation.

Replace all occurrences of $x$ in $y = x^{2} - 10 x + 12$ with $7$ .

$y = {(7)}^{2} - 10 \cdot 7 + 12$

$x = 7$

Simplify the right side.

Simplify ${(7)}^{2} - 10 \cdot 7 + 12$ .

Simplify each term.

Raise $7$ to the power of $2$ .

$y = 49 - 10 \cdot 7 + 12$

$x = 7$

Multiply $- 10$ by $7$ .

$y = 49 - 70 + 12$

$x = 7$

$y = 49 - 70 + 12$

$x = 7$

Simplify by adding and subtracting.

Subtract $70$ from $49$ .

$y = - 21 + 12$

$x = 7$

Add $- 21$ and $12$ .

$y = - 9$

$x = 7$

$y = - 9$

$x = 7$

$y = - 9$

$x = 7$

$y = - 9$

$x = 7$

$y = - 9$

$x = 7$

Step 4

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

Replace all occurrences of $x$ in $y = x^{2} - 10 x + 12$ with $1$ .

$y = {(1)}^{2} - 10 \cdot 1 + 12$

$x = 1$

Simplify the right side.

Simplify ${(1)}^{2} - 10 \cdot 1 + 12$ .

Simplify each term.

One to any power is one.

$y = 1 - 10 \cdot 1 + 12$

$x = 1$

Multiply $- 10$ by $1$ .

$y = 1 - 10 + 12$

$x = 1$

$y = 1 - 10 + 12$

$x = 1$

Simplify by adding and subtracting.

Subtract $10$ from $1$ .

$y = - 9 + 12$

$x = 1$

Add $- 9$ and $12$ .

$y = 3$

$x = 1$

$y = 3$

$x = 1$

$y = 3$

$x = 1$

$y = 3$

$x = 1$

$y = 3$

$x = 1$

Step 5

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

$(7, - 9)$

$(1, 3)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(7, - 9), (1, 3)$

Equation Form:

$x = 7, y = - 9$

$x = 1, y = 3$

Step 7

Solve the System of Equations y=x^2-10x+12 2x+y=5

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