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

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

Step 1

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

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

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

$y = x - 3$

Simplify the left side.

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

Rewrite ${(x - 3)}^{2}$ as $(x - 3) (x - 3)$ .

$(x - 3) (x - 3) = x - 3$

$y = x - 3$

Expand $(x - 3) (x - 3)$ using the FOIL Method.

Apply the distributive property.

$x (x - 3) - 3 (x - 3) = x - 3$

$y = x - 3$

Apply the distributive property.

$x \cdot x + x \cdot - 3 - 3 (x - 3) = x - 3$

$y = x - 3$

Apply the distributive property.

$x \cdot x + x \cdot - 3 - 3 x - 3 \cdot - 3 = x - 3$

$y = x - 3$

$x \cdot x + x \cdot - 3 - 3 x - 3 \cdot - 3 = x - 3$

$y = x - 3$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

$y = x - 3$

Move $- 3$ to the left of $x$ .

$x^{2} - 3 \cdot x - 3 x - 3 \cdot - 3 = x - 3$

$y = x - 3$

Multiply $- 3$ by $- 3$ .

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

$y = x - 3$

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

$y = x - 3$

Subtract $3 x$ from $- 3 x$ .

$x^{2} - 6 x + 9 = x - 3$

$y = x - 3$

$x^{2} - 6 x + 9 = x - 3$

$y = x - 3$

$x^{2} - 6 x + 9 = x - 3$

$y = x - 3$

$x^{2} - 6 x + 9 = x - 3$

$y = x - 3$

$x^{2} - 6 x + 9 = x - 3$

$y = x - 3$

Step 2

Solve for $x$ in $x^{2} - 6 x + 9 = x - 3$ .

Move all terms containing $x$ to the left side of the equation.

Subtract $x$ from both sides of the equation.

$x^{2} - 6 x + 9 - x = - 3$

$y = x - 3$

Subtract $x$ from $- 6 x$ .

$x^{2} - 7 x + 9 = - 3$

$y = x - 3$

$x^{2} - 7 x + 9 = - 3$

$y = x - 3$

Add $3$ to both sides of the equation.

$x^{2} - 7 x + 9 + 3 = 0$

$y = x - 3$

Add $9$ and $3$ .

$x^{2} - 7 x + 12 = 0$

$y = x - 3$

Factor $x^{2} - 7 x + 12$ 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 $12$ and whose sum is $- 7$ .

$- 4, - 3$

$y = x - 3$

Write the factored form using these integers.

$(x - 4) (x - 3) = 0$

$y = x - 3$

$(x - 4) (x - 3) = 0$

$y = x - 3$

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

$x - 4 = 0$

$x - 3 = 0$

$y = x - 3$

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

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

$x - 4 = 0$

$y = x - 3$

Add $4$ to both sides of the equation.

$x = 4$

$y = x - 3$

$x = 4$

$y = x - 3$

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

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

$x - 3 = 0$

$y = x - 3$

Add $3$ to both sides of the equation.

$x = 3$

$y = x - 3$

$x = 3$

$y = x - 3$

The final solution is all the values that make $(x - 4) (x - 3) = 0$ true.

$x = 4, 3$

$y = x - 3$

$x = 4, 3$

$y = x - 3$

Step 3

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

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

$y = (4) - 3$

$x = 4$

Simplify the right side.

Subtract $3$ from $4$ .

$y = 1$

$x = 4$

$y = 1$

$x = 4$

$y = 1$

$x = 4$

Step 4

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

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

$y = (3) - 3$

$x = 3$

Simplify the right side.

Subtract $3$ from $3$ .

$y = 0$

$x = 3$

$y = 0$

$x = 3$

$y = 0$

$x = 3$

Step 5

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

$(4, 1)$

$(3, 0)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(4, 1), (3, 0)$

Equation Form:

$x = 4, y = 1$

$x = 3, y = 0$

Step 7

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

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