Solve by Substitution y=x^2+2x-1 y-3x=5

$y = x^{2} + 2 x - 1$ $y - 3 x = 5$

Step 1

Replace all occurrences of $y$ with $x^{2} + 2 x - 1$ in each equation.

Replace all occurrences of $y$ in $y - 3 x = 5$ with $x^{2} + 2 x - 1$ .

$(x^{2} + 2 x - 1) - 3 x = 5$

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

Simplify the left side.

Subtract $3 x$ from $2 x$ .

$x^{2} - x - 1 = 5$

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

$x^{2} - x - 1 = 5$

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

$x^{2} - x - 1 = 5$

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

Step 2

Solve for $x$ in $x^{2} - x - 1 = 5$ .

Subtract $5$ from both sides of the equation.

$x^{2} - x - 1 - 5 = 0$

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

Subtract $5$ from $- 1$ .

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

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

Factor $x^{2} - x - 6$ 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 $- 1$ .

$- 3, 2$

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

Write the factored form using these integers.

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

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

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

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

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

$x - 3 = 0$

$x + 2 = 0$

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

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

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

$x - 3 = 0$

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

Add $3$ to both sides of the equation.

$x = 3$

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

$x = 3$

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

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

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

$x + 2 = 0$

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

Subtract $2$ from both sides of the equation.

$x = - 2$

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

$x = - 2$

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

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

$x = 3, - 2$

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

$x = 3, - 2$

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

Step 3

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

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

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

$x = 3$

Simplify the right side.

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

Simplify each term.

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

$y = 9 + 2 (3) - 1$

$x = 3$

Multiply $2$ by $3$ .

$y = 9 + 6 - 1$

$x = 3$

$y = 9 + 6 - 1$

$x = 3$

Simplify by adding and subtracting.

Add $9$ and $6$ .

$y = 15 - 1$

$x = 3$

Subtract $1$ from $15$ .

$y = 14$

$x = 3$

$y = 14$

$x = 3$

$y = 14$

$x = 3$

$y = 14$

$x = 3$

$y = 14$

$x = 3$

Step 4

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

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

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

$x = - 2$

Simplify the right side.

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

Simplify each term.

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

$y = 4 + 2 (- 2) - 1$

$x = - 2$

Multiply $2$ by $- 2$ .

$y = 4 - 4 - 1$

$x = - 2$

$y = 4 - 4 - 1$

$x = - 2$

Simplify by subtracting numbers.

Subtract $4$ from $4$ .

$y = 0 - 1$

$x = - 2$

Subtract $1$ from $0$ .

$y = - 1$

$x = - 2$

$y = - 1$

$x = - 2$

$y = - 1$

$x = - 2$

$y = - 1$

$x = - 2$

$y = - 1$

$x = - 2$

Step 5

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

$(3, 14)$

$(- 2, - 1)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(3, 14), (- 2, - 1)$

Equation Form:

$x = 3, y = 14$

$x = - 2, y = - 1$

Step 7

Solve by Substitution y=x^2+2x-1 y-3x=5

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