Solve the System of Equations y = square root of x-5 y=-x+1

$y = \sqrt{x} - 5$ $y = - x + 1$

Step 1

Eliminate the equal sides of each equation and combine.

$\sqrt{x} - 5 = - x + 1$

Step 2

Solve $\sqrt{x} - 5 = - x + 1$ for $x$ .

Move all terms not containing $\sqrt{x}$ to the right side of the equation.

Add $5$ to both sides of the equation.

$\sqrt{x} = - x + 1 + 5$

Add $1$ and $5$ .

$\sqrt{x} = - x + 6$

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{x}}^{2} = {(- x + 6)}^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

${(x^{\frac{1}{2}})}^{2} = {(- x + 6)}^{2}$

Simplify the left side.

Simplify ${(x^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${(x^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x^{\frac{1}{2} \cdot 2} = {(- x + 6)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$x^{\frac{1}{2} \cdot 2} = {(- x + 6)}^{2}$

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

Rewrite ${(- x + 6)}^{2}$ as $(- x + 6) (- x + 6)$ .

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

Expand $(- x + 6) (- x + 6)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

$x = - x (- x) - x \cdot 6 + 6 (- x + 6)$

Apply the distributive property.

$x = - x (- x) - x \cdot 6 + 6 (- x) + 6 \cdot 6$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$x = - 1 \cdot - 1 x \cdot x - x \cdot 6 + 6 (- x) + 6 \cdot 6$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$x = - 1 \cdot - 1 (x \cdot x) - x \cdot 6 + 6 (- x) + 6 \cdot 6$

Multiply $x$ by $x$ .

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

Multiply $- 1$ by $- 1$ .

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

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

$x = x^{2} - x \cdot 6 + 6 (- x) + 6 \cdot 6$

Multiply $6$ by $- 1$ .

$x = x^{2} - 6 x + 6 (- x) + 6 \cdot 6$

Multiply $- 1$ by $6$ .

$x = x^{2} - 6 x - 6 x + 6 \cdot 6$

Multiply $6$ by $6$ .

$x = x^{2} - 6 x - 6 x + 36$

Subtract $6 x$ from $- 6 x$ .

$x = x^{2} - 12 x + 36$

Solve for $x$ .

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

$x^{2} - 12 x + 36 = x$

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

Subtract $x$ from both sides of the equation.

$x^{2} - 12 x + 36 - x = 0$

Subtract $x$ from $- 12 x$ .

$x^{2} - 13 x + 36 = 0$

Factor $x^{2} - 13 x + 36$ 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 $36$ and whose sum is $- 13$ .

$- 9, - 4 + y = - x + 1$

Write the factored form using these integers.

$(x - 9) (x - 4) = 0$

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

$x - 9 = 0$

$x - 4 = 0 + y = - x + 1$

Set $x - 9$ equal to $0$ and solve for $x$ .

Set $x - 9$ equal to $0$ .

$x - 9 = 0$

Add $9$ to both sides of the equation.

$x = 9$

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

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

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

The final solution is all the values that make $(x - 9) (x - 4) = 0$ true.

$x = 9, 4$

Exclude the solutions that do not make $\sqrt{x} - 5 = - x + 1$ true.

$x = 4$

Step 3

Evaluate $y$ when $x = 4$ .

Substitute $4$ for $x$ .

$y = - (4) + 1$

Simplify $- (4) + 1$ .

Multiply $- 1$ by $4$ .

$y = - 4 + 1$

Add $- 4$ and $1$ .

$y = - 3$

Step 4

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

$(4, - 3)$

Step 5

The result can be shown in multiple forms.

Point Form:

$(4, - 3)$

Equation Form:

$x = 4, y = - 3$

Step 6

Solve the System of Equations y = square root of x-5 y=-x+1

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