Solve the System of Equations y=-x^2+5x+1 2x+y=7

$y = - x^{2} + 5 x + 1$ $2 x + y = 7$

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

Replace all occurrences of $y$ in $2 x + y = 7$ with $- x^{2} + 5 x + 1$ .

$2 x - x^{2} + 5 x + 1 = 7$

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

Simplify the left side.

Simplify $2 x - x^{2} + 5 x + 1$ .

Remove parentheses.

$2 x - x^{2} + 5 x + 1 = 7$

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

Add $2 x$ and $5 x$ .

$- x^{2} + 7 x + 1 = 7$

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

$- x^{2} + 7 x + 1 = 7$

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

$- x^{2} + 7 x + 1 = 7$

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

$- x^{2} + 7 x + 1 = 7$

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

Solve for $x$ in $- x^{2} + 7 x + 1 = 7$ .

Subtract $7$ from both sides of the equation.

$- x^{2} + 7 x + 1 - 7 = 0$

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

Subtract $7$ from $1$ .

$- x^{2} + 7 x - 6 = 0$

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

Factor the left side of the equation.

Factor $- 1$ out of $- x^{2} + 7 x - 6$ .

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

$- (x^{2}) + 7 x - 6 = 0$

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

Factor $- 1$ out of $7 x$ .

$- (x^{2}) - (- 7 x) - 6 = 0$

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

Rewrite $- 6$ as $- 1 (6)$ .

$- (x^{2}) - (- 7 x) - 1 \cdot 6 = 0$

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

Factor $- 1$ out of $- (x^{2}) - (- 7 x)$ .

$- (x^{2} - 7 x) - 1 \cdot 6 = 0$

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

Factor $- 1$ out of $- (x^{2} - 7 x) - 1 (6)$ .

$- (x^{2} - 7 x + 6) = 0$

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

$- (x^{2} - 7 x + 6) = 0$

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

Factor.

Factor $x^{2} - 7 x + 6$ 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 $- 7$ .

$- 6, - 1$

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

Write the factored form using these integers.

$- ((x - 6) (x - 1)) = 0$

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

$- ((x - 6) (x - 1)) = 0$

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

Remove unnecessary parentheses.

$- (x - 6) (x - 1) = 0$

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

$- (x - 6) (x - 1) = 0$

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

$- (x - 6) (x - 1) = 0$

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

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

$x - 6 = 0$

$x - 1 = 0$

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

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

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

Add $6$ to both sides of the equation.

$x = 6$

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

$x = 6$

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

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

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

$x - 1 = 0$

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

Add $1$ to both sides of the equation.

$x = 1$

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

$x = 1$

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

The final solution is all the values that make $- (x - 6) (x - 1) = 0$ true.

$x = 6, 1$

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

$x = 6, 1$

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

Replace all occurrences of $x$ with $6$ in each equation.

Replace all occurrences of $x$ in $y = - x^{2} + 5 x + 1$ with $6$ .

$y = - {(6)}^{2} + 5 (6) + 1$

$x = 6$

Simplify the right side.

Simplify $- {(6)}^{2} + 5 (6) + 1$ .

Simplify each term.

Raise $6$ to the power of $2$ .

$y = - 1 \cdot 36 + 5 (6) + 1$

$x = 6$

Multiply $- 1$ by $36$ .

$y = - 36 + 5 (6) + 1$

$x = 6$

Multiply $5$ by $6$ .

$y = - 36 + 30 + 1$

$x = 6$

$y = - 36 + 30 + 1$

$x = 6$

Simplify by adding numbers.

Add $- 36$ and $30$ .

$y = - 6 + 1$

$x = 6$

Add $- 6$ and $1$ .

$y = - 5$

$x = 6$

$y = - 5$

$x = 6$

$y = - 5$

$x = 6$

$y = - 5$

$x = 6$

$y = - 5$

$x = 6$

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

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

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

$x = 1$

Simplify the right side.

Simplify $- {(1)}^{2} + 5 (1) + 1$ .

Simplify each term.

One to any power is one.

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

$x = 1$

Multiply $- 1$ by $1$ .

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

$x = 1$

Multiply $5$ by $1$ .

$y = - 1 + 5 + 1$

$x = 1$

$y = - 1 + 5 + 1$

$x = 1$

Simplify by adding numbers.

Add $- 1$ and $5$ .

$y = 4 + 1$

$x = 1$

Add $4$ and $1$ .

$y = 5$

$x = 1$

$y = 5$

$x = 1$

$y = 5$

$x = 1$

$y = 5$

$x = 1$

$y = 5$

$x = 1$

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

$(6, - 5)$

$(1, 5)$

The result can be shown in multiple forms.

Point Form:

$(6, - 5), (1, 5)$

Equation Form:

$x = 6, y = - 5$

$x = 1, y = 5$

Solve the System of Equations y=-x^2+5x+1 2x+y=7

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