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

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

Step 1

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

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

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

$y = - 3 x - 3$

Simplify the right side.

Remove parentheses.

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

$y = - 3 x - 3$

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

$y = - 3 x - 3$

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

$y = - 3 x - 3$

Step 2

Solve for $x$ in $2 x - 3 = - 3 x - 3 + 5 x^{2}$ .

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

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

$y = - 3 x - 3$

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

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

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

$y = - 3 x - 3$

Subtract $2 x$ from $- 3 x$ .

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

$y = - 3 x - 3$

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

$y = - 3 x - 3$

Add $3$ to both sides of the equation.

$- 5 x - 3 + 5 x^{2} + 3 = 0$

$y = - 3 x - 3$

Combine the opposite terms in $- 5 x - 3 + 5 x^{2} + 3$ .

Add $- 3$ and $3$ .

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

$y = - 3 x - 3$

Add $- 5 x + 5 x^{2}$ and $0$ .

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

$y = - 3 x - 3$

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

$y = - 3 x - 3$

Factor $- 5 x$ out of $- 5 x + 5 x^{2}$ .

Reorder $- 5 x$ and $5 x^{2}$ .

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

$y = - 3 x - 3$

Factor $- 5 x$ out of $5 x^{2}$ .

$- 5 x (- x) - 5 x = 0$

$y = - 3 x - 3$

Factor $- 5 x$ out of $- 5 x$ .

$- 5 x (- x) - 5 x \cdot 1 = 0$

$y = - 3 x - 3$

Factor $- 5 x$ out of $- 5 x (- x) - 5 x (1)$ .

$- 5 x (- x + 1) = 0$

$y = - 3 x - 3$

$- 5 x (- x + 1) = 0$

$y = - 3 x - 3$

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

$x = 0$

$- x + 1 = 0$

$y = - 3 x - 3$

Set $x$ equal to $0$ .

$x = 0$

$y = - 3 x - 3$

Set $- x + 1$ equal to $0$ and solve for $x$ .

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

$- x + 1 = 0$

$y = - 3 x - 3$

Solve $- x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$- x = - 1$

$y = - 3 x - 3$

Divide each term in $- x = - 1$ by $- 1$ and simplify.

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

$y = - 3 x - 3$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

$y = - 3 x - 3$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

$y = - 3 x - 3$

$x = \frac{- 1}{- 1}$

$y = - 3 x - 3$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

$y = - 3 x - 3$

$x = 1$

$y = - 3 x - 3$

$x = 1$

$y = - 3 x - 3$

$x = 1$

$y = - 3 x - 3$

$x = 1$

$y = - 3 x - 3$

The final solution is all the values that make $- 5 x (- x + 1) = 0$ true.

$x = 0, 1$

$y = - 3 x - 3$

$x = 0, 1$

$y = - 3 x - 3$

Step 3

Replace all occurrences of $x$ with $0$ in each equation.

Replace all occurrences of $x$ in $y = - 3 x - 3$ with $0$ .

$y = - 3 \cdot 0 - 3$

$x = 0$

Simplify the right side.

Simplify $- 3 \cdot 0 - 3$ .

Multiply $- 3$ by $0$ .

$y = 0 - 3$

$x = 0$

Subtract $3$ from $0$ .

$y = - 3$

$x = 0$

$y = - 3$

$x = 0$

$y = - 3$

$x = 0$

$y = - 3$

$x = 0$

Step 4

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

Replace all occurrences of $x$ in $y = - 3 x - 3$ with $1$ .

$y = - 3 \cdot 1 - 3$

$x = 1$

Simplify the right side.

Simplify $- 3 \cdot 1 - 3$ .

Multiply $- 3$ by $1$ .

$y = - 3 - 3$

$x = 1$

Subtract $3$ from $- 3$ .

$y = - 6$

$x = 1$

$y = - 6$

$x = 1$

$y = - 6$

$x = 1$

$y = - 6$

$x = 1$

Step 5

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

$(0, - 3)$

$(1, - 6)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(0, - 3), (1, - 6)$

Equation Form:

$x = 0, y = - 3$

$x = 1, y = - 6$

Step 7

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

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