Solve the System of Equations y=-x^2+49 2x+y=1

$y = - x^{2} + 49$ $2 x + y = 1$

Step 1

Replace all occurrences of $y$ with $- x^{2} + 49$ in each equation.

Replace all occurrences of $y$ in $2 x + y = 1$ with $- x^{2} + 49$ .

$2 x - x^{2} + 49 = 1$

$y = - x^{2} + 49$

Simplify the left side.

Remove parentheses.

$2 x - x^{2} + 49 = 1$

$y = - x^{2} + 49$

$2 x - x^{2} + 49 = 1$

$y = - x^{2} + 49$

$2 x - x^{2} + 49 = 1$

$y = - x^{2} + 49$

Step 2

Solve for $x$ in $2 x - x^{2} + 49 = 1$ .

Subtract $1$ from both sides of the equation.

$2 x - x^{2} + 49 - 1 = 0$

$y = - x^{2} + 49$

Subtract $1$ from $49$ .

$2 x - x^{2} + 48 = 0$

$y = - x^{2} + 49$

Factor the left side of the equation.

Factor $- 1$ out of $2 x - x^{2} + 48$ .

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

$- x^{2} + 2 x + 48 = 0$

$y = - x^{2} + 49$

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

$- (x^{2}) + 2 x + 48 = 0$

$y = - x^{2} + 49$

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

$- (x^{2}) - (- 2 x) + 48 = 0$

$y = - x^{2} + 49$

Rewrite $48$ as $- 1 (- 48)$ .

$- (x^{2}) - (- 2 x) - 1 \cdot - 48 = 0$

$y = - x^{2} + 49$

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

$- (x^{2} - 2 x) - 1 \cdot - 48 = 0$

$y = - x^{2} + 49$

Factor $- 1$ out of $- (x^{2} - 2 x) - 1 (- 48)$ .

$- (x^{2} - 2 x - 48) = 0$

$y = - x^{2} + 49$

$- (x^{2} - 2 x - 48) = 0$

$y = - x^{2} + 49$

Factor.

Factor $x^{2} - 2 x - 48$ 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 $- 48$ and whose sum is $- 2$ .

$- 8, 6$

$y = - x^{2} + 49$

Write the factored form using these integers.

$- ((x - 8) (x + 6)) = 0$

$y = - x^{2} + 49$

$- ((x - 8) (x + 6)) = 0$

$y = - x^{2} + 49$

Remove unnecessary parentheses.

$- (x - 8) (x + 6) = 0$

$y = - x^{2} + 49$

$- (x - 8) (x + 6) = 0$

$y = - x^{2} + 49$

$- (x - 8) (x + 6) = 0$

$y = - x^{2} + 49$

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

$x - 8 = 0$

$x + 6 = 0$

$y = - x^{2} + 49$

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

$y = - x^{2} + 49$

Add $8$ to both sides of the equation.

$x = 8$

$y = - x^{2} + 49$

$x = 8$

$y = - x^{2} + 49$

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

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

$x + 6 = 0$

$y = - x^{2} + 49$

Subtract $6$ from both sides of the equation.

$x = - 6$

$y = - x^{2} + 49$

$x = - 6$

$y = - x^{2} + 49$

The final solution is all the values that make $- (x - 8) (x + 6) = 0$ true.

$x = 8, - 6$

$y = - x^{2} + 49$

$x = 8, - 6$

$y = - x^{2} + 49$

Step 3

Replace all occurrences of $x$ with $8$ in each equation.

Replace all occurrences of $x$ in $y = - x^{2} + 49$ with $8$ .

$y = - {(8)}^{2} + 49$

$x = 8$

Simplify the right side.

Simplify $- {(8)}^{2} + 49$ .

Simplify each term.

Raise $8$ to the power of $2$ .

$y = - 1 \cdot 64 + 49$

$x = 8$

Multiply $- 1$ by $64$ .

$y = - 64 + 49$

$x = 8$

$y = - 64 + 49$

$x = 8$

Add $- 64$ and $49$ .

$y = - 15$

$x = 8$

$y = - 15$

$x = 8$

$y = - 15$

$x = 8$

$y = - 15$

$x = 8$

Step 4

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

Replace all occurrences of $x$ in $y = - x^{2} + 49$ with $- 6$ .

$y = - {(- 6)}^{2} + 49$

$x = - 6$

Simplify the right side.

Simplify $- {(- 6)}^{2} + 49$ .

Simplify each term.

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

$y = - 1 \cdot 36 + 49$

$x = - 6$

Multiply $- 1$ by $36$ .

$y = - 36 + 49$

$x = - 6$

$y = - 36 + 49$

$x = - 6$

Add $- 36$ and $49$ .

$y = 13$

$x = - 6$

$y = 13$

$x = - 6$

$y = 13$

$x = - 6$

$y = 13$

$x = - 6$

Step 5

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

$(8, - 15)$

$(- 6, 13)$

Step 6

The result can be shown in multiple forms.

Point Form:

$(8, - 15), (- 6, 13)$

Equation Form:

$x = 8, y = - 15$

$x = - 6, y = 13$

Step 7

Solve the System of Equations y=-x^2+49 2x+y=1

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