Solve by Substitution x^2-y=0 , 2x+y=8

$x^{2} - y = 0$ , $2 x + y = 8$

Step 1

Subtract $2 x$ from both sides of the equation.

$y = 8 - 2 x$

$x^{2} - y = 0$

Step 2

Replace all occurrences of $y$ with $8 - 2 x$ in each equation.

Replace all occurrences of $y$ in $x^{2} - y = 0$ with $8 - 2 x$ .

$x^{2} - (8 - 2 x) = 0$

$y = 8 - 2 x$

Simplify the left side.

Simplify each term.

Apply the distributive property.

$x^{2} - 1 \cdot 8 - (- 2 x) = 0$

$y = 8 - 2 x$

Multiply $- 1$ by $8$ .

$x^{2} - 8 - (- 2 x) = 0$

$y = 8 - 2 x$

Multiply $- 2$ by $- 1$ .

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

$y = 8 - 2 x$

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

$y = 8 - 2 x$

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

$y = 8 - 2 x$

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

$y = 8 - 2 x$

Step 3

Solve for $x$ in $x^{2} - 8 + 2 x = 0$ .

Factor $x^{2} - 8 + 2 x$ 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 $2$ .

$- 2, 4$

$y = 8 - 2 x$

Write the factored form using these integers.

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

$y = 8 - 2 x$

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

$y = 8 - 2 x$

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

$x - 2 = 0$

$x + 4 = 0$

$y = 8 - 2 x$

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

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

$x - 2 = 0$

$y = 8 - 2 x$

Add $2$ to both sides of the equation.

$x = 2$

$y = 8 - 2 x$

$x = 2$

$y = 8 - 2 x$

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

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

$x + 4 = 0$

$y = 8 - 2 x$

Subtract $4$ from both sides of the equation.

$x = - 4$

$y = 8 - 2 x$

$x = - 4$

$y = 8 - 2 x$

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

$x = 2, - 4$

$y = 8 - 2 x$

$x = 2, - 4$

$y = 8 - 2 x$

Step 4

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

Replace all occurrences of $x$ in $y = 8 - 2 x$ with $2$ .

$y = 8 - 2 \cdot 2$

$x = 2$

Simplify the right side.

Simplify $8 - 2 \cdot 2$ .

Multiply $- 2$ by $2$ .

$y = 8 - 4$

$x = 2$

Subtract $4$ from $8$ .

$y = 4$

$x = 2$

$y = 4$

$x = 2$

$y = 4$

$x = 2$

$y = 4$

$x = 2$

Step 5

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

Replace all occurrences of $x$ in $y = 8 - 2 x$ with $- 4$ .

$y = 8 - 2 \cdot - 4$

$x = - 4$

Simplify the right side.

Simplify $8 - 2 \cdot - 4$ .

Multiply $- 2$ by $- 4$ .

$y = 8 + 8$

$x = - 4$

Add $8$ and $8$ .

$y = 16$

$x = - 4$

$y = 16$

$x = - 4$

$y = 16$

$x = - 4$

$y = 16$

$x = - 4$

Step 6

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

$(2, 4)$

$(- 4, 16)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(2, 4), (- 4, 16)$

Equation Form:

$x = 2, y = 4$

$x = - 4, y = 16$

Step 8

Solve by Substitution x^2-y=0 , 2x+y=8

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