Solve by Substitution y=x^2-3 , 2x-y=4

$y = x^{2} - 3$ , $2 x - y = 4$

Step 1

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

Replace all occurrences of $y$ in $2 x - y = 4$ with $x^{2} - 3$ .

$2 x - (x^{2} - 3) = 4$

$y = x^{2} - 3$

Simplify the left side.

Simplify each term.

Apply the distributive property.

$2 x - x^{2} + 3 = 4$

$y = x^{2} - 3$

Multiply $- 1$ by $- 3$ .

$2 x - x^{2} + 3 = 4$

$y = x^{2} - 3$

$2 x - x^{2} + 3 = 4$

$y = x^{2} - 3$

$2 x - x^{2} + 3 = 4$

$y = x^{2} - 3$

$2 x - x^{2} + 3 = 4$

$y = x^{2} - 3$

Step 2

Solve for $x$ in $2 x - x^{2} + 3 = 4$ .

Subtract $4$ from both sides of the equation.

$2 x - x^{2} + 3 - 4 = 0$

$y = x^{2} - 3$

Subtract $4$ from $3$ .

$2 x - x^{2} - 1 = 0$

$y = x^{2} - 3$

Factor the left side of the equation.

Factor $- 1$ out of $2 x - x^{2} - 1$ .

Reorder $2 x$ and $- x^{2}$ .

$- x^{2} + 2 x - 1 = 0$

$y = x^{2} - 3$

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 2 x - 1 = 0$

$y = x^{2} - 3$

Factor $- 1$ out of $2 x$ .

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

$y = x^{2} - 3$

Rewrite $- 1$ as $- 1 (1)$ .

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

$y = x^{2} - 3$

Factor $- 1$ out of $- (x^{2}) - (- 2 x)$ .

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

$y = x^{2} - 3$

Factor $- 1$ out of $- (x^{2} - 2 x) - 1 (1)$ .

$- (x^{2} - 2 x + 1) = 0$

$y = x^{2} - 3$

$- (x^{2} - 2 x + 1) = 0$

$y = x^{2} - 3$

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

$- (x^{2} - 2 x + 1^{2}) = 0$

$y = x^{2} - 3$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

$y = x^{2} - 3$

Rewrite the polynomial.

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

$y = x^{2} - 3$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 1$ .

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

$y = x^{2} - 3$

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

$y = x^{2} - 3$

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

$y = x^{2} - 3$

Divide each term in $- {(x - 1)}^{2} = 0$ by $- 1$ and simplify.

Divide each term in $- {(x - 1)}^{2} = 0$ by $- 1$ .

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

$y = x^{2} - 3$

Simplify the left side.

Dividing two negative values results in a positive value.

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

$y = x^{2} - 3$

Divide ${(x - 1)}^{2}$ by $1$ .

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

$y = x^{2} - 3$

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

$y = x^{2} - 3$

Simplify the right side.

Divide $0$ by $- 1$ .

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

$y = x^{2} - 3$

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

$y = x^{2} - 3$

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

$y = x^{2} - 3$

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

$x - 1 = 0$

$y = x^{2} - 3$

Add $1$ to both sides of the equation.

$x = 1$

$y = x^{2} - 3$

$x = 1$

$y = x^{2} - 3$

Step 3

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

Replace all occurrences of $x$ in $y = x^{2} - 3$ with $1$ .

$y = {(1)}^{2} - 3$

$x = 1$

Simplify the right side.

Simplify ${(1)}^{2} - 3$ .

One to any power is one.

$y = 1 - 3$

$x = 1$

Subtract $3$ from $1$ .

$y = - 2$

$x = 1$

$y = - 2$

$x = 1$

$y = - 2$

$x = 1$

$y = - 2$

$x = 1$

Step 4

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

$(1, - 2)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(1, - 2)$

Equation Form:

$x = 1, y = - 2$

Step 6

Solve by Substitution y=x^2-3 , 2x-y=4

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