Solve the System of Equations x^2+2y=10 y=-3x+9

$x^{2} + 2 y = 10$ $y = - 3 x + 9$

Step 1

Replace all occurrences of $y$ with $- 3 x + 9$ in each equation.

Replace all occurrences of $y$ in $x^{2} + 2 y = 10$ with $- 3 x + 9$ .

$x^{2} + 2 (- 3 x + 9) = 10$

$y = - 3 x + 9$

Simplify the left side.

Simplify each term.

Apply the distributive property.

$x^{2} + 2 (- 3 x) + 2 \cdot 9 = 10$

$y = - 3 x + 9$

Multiply $- 3$ by $2$ .

$x^{2} - 6 x + 2 \cdot 9 = 10$

$y = - 3 x + 9$

Multiply $2$ by $9$ .

$x^{2} - 6 x + 18 = 10$

$y = - 3 x + 9$

$x^{2} - 6 x + 18 = 10$

$y = - 3 x + 9$

$x^{2} - 6 x + 18 = 10$

$y = - 3 x + 9$

$x^{2} - 6 x + 18 = 10$

$y = - 3 x + 9$

Step 2

Solve for $x$ in $x^{2} - 6 x + 18 = 10$ .

Subtract $10$ from both sides of the equation.

$x^{2} - 6 x + 18 - 10 = 0$

$y = - 3 x + 9$

Subtract $10$ from $18$ .

$x^{2} - 6 x + 8 = 0$

$y = - 3 x + 9$

Factor $x^{2} - 6 x + 8$ 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 $8$ and whose sum is $- 6$ .

$- 4, - 2$

$y = - 3 x + 9$

Write the factored form using these integers.

$(x - 4) (x - 2) = 0$

$y = - 3 x + 9$

$(x - 4) (x - 2) = 0$

$y = - 3 x + 9$

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

$x - 4 = 0$

$x - 2 = 0$

$y = - 3 x + 9$

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

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

$x - 4 = 0$

$y = - 3 x + 9$

Add $4$ to both sides of the equation.

$x = 4$

$y = - 3 x + 9$

$x = 4$

$y = - 3 x + 9$

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

$y = - 3 x + 9$

Add $2$ to both sides of the equation.

$x = 2$

$y = - 3 x + 9$

$x = 2$

$y = - 3 x + 9$

The final solution is all the values that make $(x - 4) (x - 2) = 0$ true.

$x = 4, 2$

$y = - 3 x + 9$

$x = 4, 2$

$y = - 3 x + 9$

Step 3

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

Replace all occurrences of $x$ in $y = - 3 x + 9$ with $4$ .

$y = - 3 \cdot 4 + 9$

$x = 4$

Simplify the right side.

Simplify $- 3 \cdot 4 + 9$ .

Multiply $- 3$ by $4$ .

$y = - 12 + 9$

$x = 4$

Add $- 12$ and $9$ .

$y = - 3$

$x = 4$

$y = - 3$

$x = 4$

$y = - 3$

$x = 4$

$y = - 3$

$x = 4$

Step 4

Replace all occurrences of $x$ with $2$ in each equation.

Replace all occurrences of $x$ in $y = - 3 x + 9$ with $2$ .

$y = - 3 \cdot 2 + 9$

$x = 2$

Simplify the right side.

Simplify $- 3 \cdot 2 + 9$ .

Multiply $- 3$ by $2$ .

$y = - 6 + 9$

$x = 2$

Add $- 6$ and $9$ .

$y = 3$

$x = 2$

$y = 3$

$x = 2$

$y = 3$

$x = 2$

$y = 3$

$x = 2$

Step 5

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

$(4, - 3)$

$(2, 3)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, - 3), (2, 3)$

Equation Form:

$x = 4, y = - 3$

$x = 2, y = 3$

Step 7

Solve the System of Equations x^2+2y=10 y=-3x+9

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