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

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

Step 1

Subtract $y$ from both sides of the equation.

$x = 5 - y$

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

Step 2

Replace all occurrences of $x$ with $5 - y$ in each equation.

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

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

$x = 5 - y$

Simplify the right side.

Simplify $3 {(5 - y)}^{2} + 2 (5 - y)$ .

Simplify each term.

Rewrite ${(5 - y)}^{2}$ as $(5 - y) (5 - y)$ .

$y + 1 = 3 ((5 - y) (5 - y)) + 2 (5 - y)$

$x = 5 - y$

Expand $(5 - y) (5 - y)$ using the FOIL Method.

Apply the distributive property.

$y + 1 = 3 (5 (5 - y) - y (5 - y)) + 2 (5 - y)$

$x = 5 - y$

Apply the distributive property.

$y + 1 = 3 (5 \cdot 5 + 5 (- y) - y (5 - y)) + 2 (5 - y)$

$x = 5 - y$

Apply the distributive property.

$y + 1 = 3 (5 \cdot 5 + 5 (- y) - y \cdot 5 - y (- y)) + 2 (5 - y)$

$x = 5 - y$

$y + 1 = 3 (5 \cdot 5 + 5 (- y) - y \cdot 5 - y (- y)) + 2 (5 - y)$

$x = 5 - y$

Simplify and combine like terms.

Simplify each term.

Multiply $5$ by $5$ .

$y + 1 = 3 (25 + 5 (- y) - y \cdot 5 - y (- y)) + 2 (5 - y)$

$x = 5 - y$

Multiply $- 1$ by $5$ .

$y + 1 = 3 (25 - 5 y - y \cdot 5 - y (- y)) + 2 (5 - y)$

$x = 5 - y$

Multiply $5$ by $- 1$ .

$y + 1 = 3 (25 - 5 y - 5 y - y (- y)) + 2 (5 - y)$

$x = 5 - y$

Rewrite using the commutative property of multiplication.

$y + 1 = 3 (25 - 5 y - 5 y - 1 \cdot (- 1 y \cdot y)) + 2 (5 - y)$

$x = 5 - y$

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$y + 1 = 3 (25 - 5 y - 5 y - 1 \cdot (- 1 (y \cdot y))) + 2 (5 - y)$

$x = 5 - y$

Multiply $y$ by $y$ .

$y + 1 = 3 (25 - 5 y - 5 y - 1 \cdot (- 1 y^{2})) + 2 (5 - y)$

$x = 5 - y$

$y + 1 = 3 (25 - 5 y - 5 y - 1 \cdot (- 1 y^{2})) + 2 (5 - y)$

$x = 5 - y$

Multiply $- 1$ by $- 1$ .

$y + 1 = 3 (25 - 5 y - 5 y + 1 y^{2}) + 2 (5 - y)$

$x = 5 - y$

Multiply $y^{2}$ by $1$ .

$y + 1 = 3 (25 - 5 y - 5 y + y^{2}) + 2 (5 - y)$

$x = 5 - y$

$y + 1 = 3 (25 - 5 y - 5 y + y^{2}) + 2 (5 - y)$

$x = 5 - y$

Subtract $5 y$ from $- 5 y$ .

$y + 1 = 3 (25 - 10 y + y^{2}) + 2 (5 - y)$

$x = 5 - y$

$y + 1 = 3 (25 - 10 y + y^{2}) + 2 (5 - y)$

$x = 5 - y$

Apply the distributive property.

$y + 1 = 3 \cdot 25 + 3 (- 10 y) + 3 y^{2} + 2 (5 - y)$

$x = 5 - y$

Simplify.

Multiply $3$ by $25$ .

$y + 1 = 75 + 3 (- 10 y) + 3 y^{2} + 2 (5 - y)$

$x = 5 - y$

Multiply $- 10$ by $3$ .

$y + 1 = 75 - 30 y + 3 y^{2} + 2 (5 - y)$

$x = 5 - y$

$y + 1 = 75 - 30 y + 3 y^{2} + 2 (5 - y)$

$x = 5 - y$

Apply the distributive property.

$y + 1 = 75 - 30 y + 3 y^{2} + 2 \cdot 5 + 2 (- y)$

$x = 5 - y$

Multiply $2$ by $5$ .

$y + 1 = 75 - 30 y + 3 y^{2} + 10 + 2 (- y)$

$x = 5 - y$

Multiply $- 1$ by $2$ .

$y + 1 = 75 - 30 y + 3 y^{2} + 10 - 2 y$

$x = 5 - y$

$y + 1 = 75 - 30 y + 3 y^{2} + 10 - 2 y$

$x = 5 - y$

Simplify by adding terms.

Add $75$ and $10$ .

$y + 1 = - 30 y + 3 y^{2} + 85 - 2 y$

$x = 5 - y$

Subtract $2 y$ from $- 30 y$ .

$y + 1 = - 32 y + 3 y^{2} + 85$

$x = 5 - y$

$y + 1 = - 32 y + 3 y^{2} + 85$

$x = 5 - y$

$y + 1 = - 32 y + 3 y^{2} + 85$

$x = 5 - y$

$y + 1 = - 32 y + 3 y^{2} + 85$

$x = 5 - y$

$y + 1 = - 32 y + 3 y^{2} + 85$

$x = 5 - y$

Step 3

Solve for $y$ in $y + 1 = - 32 y + 3 y^{2} + 85$ .

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

$- 32 y + 3 y^{2} + 85 = y + 1$

$x = 5 - y$

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

Subtract $y$ from both sides of the equation.

$- 32 y + 3 y^{2} + 85 - y = 1$

$x = 5 - y$

Subtract $y$ from $- 32 y$ .

$- 33 y + 3 y^{2} + 85 = 1$

$x = 5 - y$

$- 33 y + 3 y^{2} + 85 = 1$

$x = 5 - y$

Subtract $1$ from both sides of the equation.

$- 33 y + 3 y^{2} + 85 - 1 = 0$

$x = 5 - y$

Subtract $1$ from $85$ .

$- 33 y + 3 y^{2} + 84 = 0$

$x = 5 - y$

Factor the left side of the equation.

Factor $3$ out of $- 33 y + 3 y^{2} + 84$ .

Factor $3$ out of $- 33 y$ .

$3 (- 11 y) + 3 y^{2} + 84 = 0$

$x = 5 - y$

Factor $3$ out of $3 y^{2}$ .

$3 (- 11 y) + 3 (y^{2}) + 84 = 0$

$x = 5 - y$

Factor $3$ out of $84$ .

$3 (- 11 y) + 3 y^{2} + 3 \cdot 28 = 0$

$x = 5 - y$

Factor $3$ out of $3 (- 11 y) + 3 y^{2}$ .

$3 (- 11 y + y^{2}) + 3 \cdot 28 = 0$

$x = 5 - y$

Factor $3$ out of $3 (- 11 y + y^{2}) + 3 \cdot 28$ .

$3 (- 11 y + y^{2} + 28) = 0$

$x = 5 - y$

$3 (- 11 y + y^{2} + 28) = 0$

$x = 5 - y$

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

$3 (- 11 u + u^{2} + 28) = 0$

$x = 5 - y$

Factor $- 11 u + u^{2} + 28$ 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 $28$ and whose sum is $- 11$ .

$- 7, - 4$

$x = 5 - y$

Write the factored form using these integers.

$3 ((u - 7) (u - 4)) = 0$

$x = 5 - y$

$3 ((u - 7) (u - 4)) = 0$

$x = 5 - y$

Factor.

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

$3 ((y - 7) (y - 4)) = 0$

$x = 5 - y$

Remove unnecessary parentheses.

$3 (y - 7) (y - 4) = 0$

$x = 5 - y$

$3 (y - 7) (y - 4) = 0$

$x = 5 - y$

$3 (y - 7) (y - 4) = 0$

$x = 5 - y$

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

$y - 7 = 0$

$y - 4 = 0$

$x = 5 - y$

Set $y - 7$ equal to $0$ and solve for $y$ .

Set $y - 7$ equal to $0$ .

$y - 7 = 0$

$x = 5 - y$

Add $7$ to both sides of the equation.

$y = 7$

$x = 5 - y$

$y = 7$

$x = 5 - y$

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

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

$y - 4 = 0$

$x = 5 - y$

Add $4$ to both sides of the equation.

$y = 4$

$x = 5 - y$

$y = 4$

$x = 5 - y$

The final solution is all the values that make $3 (y - 7) (y - 4) = 0$ true.

$y = 7, 4$

$x = 5 - y$

$y = 7, 4$

$x = 5 - y$

Step 4

Replace all occurrences of $y$ with $7$ in each equation.

Replace all occurrences of $y$ in $x = 5 - y$ with $7$ .

$x = 5 - (7)$

$y = 7$

Simplify the right side.

Simplify $5 - (7)$ .

Multiply $- 1$ by $7$ .

$x = 5 - 7$

$y = 7$

Subtract $7$ from $5$ .

$x = - 2$

$y = 7$

$x = - 2$

$y = 7$

$x = - 2$

$y = 7$

$x = - 2$

$y = 7$

Step 5

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

Replace all occurrences of $y$ in $x = 5 - y$ with $4$ .

$x = 5 - (4)$

$y = 4$

Simplify the right side.

Simplify $5 - (4)$ .

Multiply $- 1$ by $4$ .

$x = 5 - 4$

$y = 4$

Subtract $4$ from $5$ .

$x = 1$

$y = 4$

$x = 1$

$y = 4$

$x = 1$

$y = 4$

$x = 1$

$y = 4$

Step 6

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

$(- 2, 7)$

$(1, 4)$

Step 7

The result can be shown in multiple forms.

Point Form:

$(- 2, 7), (1, 4)$

Equation Form:

$x = - 2, y = 7$

$x = 1, y = 4$

Step 8

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