Solve by Substitution x+y=3 , y=x^2-3

$x + y = 3$ , $y = x^{2} - 3$

Step 1

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

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

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

$y = x^{2} - 3$

Simplify the left side.

Remove parentheses.

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

$y = x^{2} - 3$

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

$y = x^{2} - 3$

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

$y = x^{2} - 3$

Step 2

Solve for $x$ in $x + x^{2} - 3 = 3$ .

Subtract $3$ from both sides of the equation.

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

$y = x^{2} - 3$

Subtract $3$ from $- 3$ .

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

$y = x^{2} - 3$

Factor the left side of the equation.

Let $u = x$ . Substitute $u$ for all occurrences of $x$ .

$u + u^{2} - 6 = 0$

$y = x^{2} - 3$

Factor $u + u^{2} - 6$ 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 $- 6$ and whose sum is $1$ .

$- 2, 3$

$y = x^{2} - 3$

Write the factored form using these integers.

$(u - 2) (u + 3) = 0$

$y = x^{2} - 3$

$(u - 2) (u + 3) = 0$

$y = x^{2} - 3$

Replace all occurrences of $u$ with $x$ .

$(x - 2) (x + 3) = 0$

$y = x^{2} - 3$

$(x - 2) (x + 3) = 0$

$y = x^{2} - 3$

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

$x - 2 = 0$

$x + 3 = 0$

$y = x^{2} - 3$

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

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

$x - 2 = 0$

$y = x^{2} - 3$

Add $2$ to both sides of the equation.

$x = 2$

$y = x^{2} - 3$

$x = 2$

$y = x^{2} - 3$

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

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

$x + 3 = 0$

$y = x^{2} - 3$

Subtract $3$ from both sides of the equation.

$x = - 3$

$y = x^{2} - 3$

$x = - 3$

$y = x^{2} - 3$

The final solution is all the values that make $(x - 2) (x + 3) = 0$ true.

$x = 2, - 3$

$y = x^{2} - 3$

$x = 2, - 3$

$y = x^{2} - 3$

Step 3

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

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

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

$x = 2$

Simplify the right side.

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

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

$y = 4 - 3$

$x = 2$

Subtract $3$ from $4$ .

$y = 1$

$x = 2$

$y = 1$

$x = 2$

$y = 1$

$x = 2$

$y = 1$

$x = 2$

Step 4

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

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

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

$x = - 3$

Simplify the right side.

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

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

$y = 9 - 3$

$x = - 3$

Subtract $3$ from $9$ .

$y = 6$

$x = - 3$

$y = 6$

$x = - 3$

$y = 6$

$x = - 3$

$y = 6$

$x = - 3$

Step 5

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

$(2, 1)$

$(- 3, 6)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(2, 1), (- 3, 6)$

Equation Form:

$x = 2, y = 1$

$x = - 3, y = 6$

Step 7

Solve by Substitution x+y=3 , y=x^2-3

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