Solve the System of Equations 4x^2+y^2=52 xy=12

$4 x^{2} + y^{2} = 52$ $x y = 12$

Step 1

Divide each term in $x y = 12$ by $y$ and simplify.

Divide each term in $x y = 12$ by $y$ .

$\frac{x y}{y} = \frac{12}{y}$

$4 x^{2} + y^{2} = 52$

Simplify the left side.

Cancel the common factor of $y$ .

Cancel the common factor.

$\frac{x y}{y} = \frac{12}{y}$

$4 x^{2} + y^{2} = 52$

Divide $x$ by $1$ .

$x = \frac{12}{y}$

$4 x^{2} + y^{2} = 52$

$x = \frac{12}{y}$

$4 x^{2} + y^{2} = 52$

$x = \frac{12}{y}$

$4 x^{2} + y^{2} = 52$

$x = \frac{12}{y}$

$4 x^{2} + y^{2} = 52$

Step 2

Replace all occurrences of $x$ with $\frac{12}{y}$ in each equation.

Replace all occurrences of $x$ in $4 x^{2} + y^{2} = 52$ with $\frac{12}{y}$ .

$4 {(\frac{12}{y})}^{2} + y^{2} = 52$

$x = \frac{12}{y}$

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{12}{y}$ .

$4 (\frac{12^{2}}{y^{2}}) + y^{2} = 52$

$x = \frac{12}{y}$

Raise $12$ to the power of $2$ .

$4 (\frac{144}{y^{2}}) + y^{2} = 52$

$x = \frac{12}{y}$

Multiply $4 \frac{144}{y^{2}}$ .

Combine $4$ and $\frac{144}{y^{2}}$ .

$\frac{4 \cdot 144}{y^{2}} + y^{2} = 52$

$x = \frac{12}{y}$

Multiply $4$ by $144$ .

$\frac{576}{y^{2}} + y^{2} = 52$

$x = \frac{12}{y}$

$\frac{576}{y^{2}} + y^{2} = 52$

$x = \frac{12}{y}$

$\frac{576}{y^{2}} + y^{2} = 52$

$x = \frac{12}{y}$

$\frac{576}{y^{2}} + y^{2} = 52$

$x = \frac{12}{y}$

$\frac{576}{y^{2}} + y^{2} = 52$

$x = \frac{12}{y}$

Step 3

Solve for $y$ in $\frac{576}{y^{2}} + y^{2} = 52$ .

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$y^{2}, 1, 1$

$x = \frac{12}{y}$

The LCM of one and any expression is the expression.

$y^{2}$

$x = \frac{12}{y}$

$y^{2}$

$x = \frac{12}{y}$

Multiply each term in $\frac{576}{y^{2}} + y^{2} = 52$ by $y^{2}$ to eliminate the fractions.

Multiply each term in $\frac{576}{y^{2}} + y^{2} = 52$ by $y^{2}$ .

$\frac{576}{y^{2}} \cdot y^{2} + y^{2} y^{2} = 52 y^{2}$

$x = \frac{12}{y}$

Simplify the left side.

Simplify each term.

Cancel the common factor of $y^{2}$ .

Cancel the common factor.

$\frac{576}{y^{2}} \cdot y^{2} + y^{2} y^{2} = 52 y^{2}$

$x = \frac{12}{y}$

Rewrite the expression.

$576 + y^{2} y^{2} = 52 y^{2}$

$x = \frac{12}{y}$

$576 + y^{2} y^{2} = 52 y^{2}$

$x = \frac{12}{y}$

Multiply $y^{2}$ by $y^{2}$ by adding the exponents.

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$576 + y^{2 + 2} = 52 y^{2}$

$x = \frac{12}{y}$

Add $2$ and $2$ .

$576 + y^{4} = 52 y^{2}$

$x = \frac{12}{y}$

$576 + y^{4} = 52 y^{2}$

$x = \frac{12}{y}$

$576 + y^{4} = 52 y^{2}$

$x = \frac{12}{y}$

$576 + y^{4} = 52 y^{2}$

$x = \frac{12}{y}$

$576 + y^{4} = 52 y^{2}$

$x = \frac{12}{y}$

Solve the equation.

Subtract $52 y^{2}$ from both sides of the equation.

$576 + y^{4} - 52 y^{2} = 0$

$x = \frac{12}{y}$

Substitute $u = y^{2}$ into the equation. This will make the quadratic formula easy to use.

$u^{2} - 52 u + 576 = 0$

$u = y^{2}$

$x = \frac{12}{y}$

Factor $u^{2} - 52 u + 576$ 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 $576$ and whose sum is $- 52$ .

$- 36, - 16$

$x = \frac{12}{y}$

Write the factored form using these integers.

$(u - 36) (u - 16) = 0$

$x = \frac{12}{y}$

$(u - 36) (u - 16) = 0$

$x = \frac{12}{y}$

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

$u - 36 = 0$

$u - 16 = 0$

$x = \frac{12}{y}$

Set $u - 36$ equal to $0$ and solve for $u$ .

Set $u - 36$ equal to $0$ .

$u - 36 = 0$

$x = \frac{12}{y}$

Add $36$ to both sides of the equation.

$u = 36$

$x = \frac{12}{y}$

$u = 36$

$x = \frac{12}{y}$

Set $u - 16$ equal to $0$ and solve for $u$ .

Set $u - 16$ equal to $0$ .

$u - 16 = 0$

$x = \frac{12}{y}$

Add $16$ to both sides of the equation.

$u = 16$

$x = \frac{12}{y}$

$u = 16$

$x = \frac{12}{y}$

The final solution is all the values that make $(u - 36) (u - 16) = 0$ true.

$u = 36, 16$

$x = \frac{12}{y}$

Substitute the real value of $u = y^{2}$ back into the solved equation.

$y^{2} = 36$

$y^{2} = 16$

$x = \frac{12}{y}$

Solve the first equation for $y$ .

$y^{2} = 36$

$x = \frac{12}{y}$

Solve the equation for $y$ .

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{36}$

$x = \frac{12}{y}$

Simplify $\pm \sqrt{36}$ .

Rewrite $36$ as $6^{2}$ .

$y = \pm \sqrt{6^{2}}$

$x = \frac{12}{y}$

Pull terms out from under the radical, assuming positive real numbers.

$y = \pm 6$

$x = \frac{12}{y}$

$y = \pm 6$

$x = \frac{12}{y}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$y = 6$

$x = \frac{12}{y}$

Next, use the negative value of the $\pm$ to find the second solution.

$y = - 6$

$x = \frac{12}{y}$

The complete solution is the result of both the positive and negative portions of the solution.

$y = 6, - 6$

$x = \frac{12}{y}$

$y = 6, - 6$

$x = \frac{12}{y}$

$y = 6, - 6$

$x = \frac{12}{y}$

Solve the second equation for $y$ .

$y^{2} = 16$

$x = \frac{12}{y}$

Solve the equation for $y$ .

Remove parentheses.

$y^{2} = 16$

$x = \frac{12}{y}$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{16}$

$x = \frac{12}{y}$

Simplify $\pm \sqrt{16}$ .

Rewrite $16$ as $4^{2}$ .

$y = \pm \sqrt{4^{2}}$

$x = \frac{12}{y}$

Pull terms out from under the radical, assuming positive real numbers.

$y = \pm 4$

$x = \frac{12}{y}$

$y = \pm 4$

$x = \frac{12}{y}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$y = 4$

$x = \frac{12}{y}$

Next, use the negative value of the $\pm$ to find the second solution.

$y = - 4$

$x = \frac{12}{y}$

The complete solution is the result of both the positive and negative portions of the solution.

$y = 4, - 4$

$x = \frac{12}{y}$

$y = 4, - 4$

$x = \frac{12}{y}$

$y = 4, - 4$

$x = \frac{12}{y}$

The solution to $y^{4} - 52 y^{2} + 576 = 0$ is $y = 6, - 6, 4, - 4$ .

$y = 6, - 6, 4, - 4$

$x = \frac{12}{y}$

$y = 6, - 6, 4, - 4$

$x = \frac{12}{y}$

$y = 6, - 6, 4, - 4$

$x = \frac{12}{y}$

Step 4

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

Replace all occurrences of $y$ in $x = \frac{12}{y}$ with $6$ .

$x = \frac{12}{6}$

$y = 6$

Simplify the right side.

Divide $12$ by $6$ .

$x = 2$

$y = 6$

$x = 2$

$y = 6$

$x = 2$

$y = 6$

Step 5

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

Replace all occurrences of $y$ in $x = \frac{12}{y}$ with $- 6$ .

$x = \frac{12}{- 6}$

$y = - 6$

Simplify the right side.

Divide $12$ by $- 6$ .

$x = - 2$

$y = - 6$

$x = - 2$

$y = - 6$

$x = - 2$

$y = - 6$

Step 6

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

Replace all occurrences of $y$ in $x = \frac{12}{y}$ with $6$ .

$x = \frac{12}{6}$

$y = 6$

Simplify the right side.

Divide $12$ by $6$ .

$x = 2$

$y = 6$

$x = 2$

$y = 6$

$x = 2$

$y = 6$

Step 7

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

Replace all occurrences of $y$ in $x = \frac{12}{y}$ with $- 6$ .

$x = \frac{12}{- 6}$

$y = - 6$

Simplify the right side.

Divide $12$ by $- 6$ .

$x = - 2$

$y = - 6$

$x = - 2$

$y = - 6$

$x = - 2$

$y = - 6$

Step 8

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

Replace all occurrences of $y$ in $x = \frac{12}{y}$ with $4$ .

$x = \frac{12}{4}$

$y = 4$

Simplify the right side.

Divide $12$ by $4$ .

$x = 3$

$y = 4$

$x = 3$

$y = 4$

$x = 3$

$y = 4$

Step 9

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

Replace all occurrences of $y$ in $x = \frac{12}{y}$ with $6$ .

$x = \frac{12}{6}$

$y = 6$

Simplify the right side.

Divide $12$ by $6$ .

$x = 2$

$y = 6$

$x = 2$

$y = 6$

$x = 2$

$y = 6$

Step 10

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

Replace all occurrences of $y$ in $x = \frac{12}{y}$ with $- 6$ .

$x = \frac{12}{- 6}$

$y = - 6$

Simplify the right side.

Divide $12$ by $- 6$ .

$x = - 2$

$y = - 6$

$x = - 2$

$y = - 6$

$x = - 2$

$y = - 6$

Step 11

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

Replace all occurrences of $y$ in $x = \frac{12}{y}$ with $4$ .

$x = \frac{12}{4}$

$y = 4$

Simplify the right side.

Divide $12$ by $4$ .

$x = 3$

$y = 4$

$x = 3$

$y = 4$

$x = 3$

$y = 4$

Step 12

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

Replace all occurrences of $y$ in $x = \frac{12}{y}$ with $- 4$ .

$x = \frac{12}{- 4}$

$y = - 4$

Simplify the right side.

Divide $12$ by $- 4$ .

$x = - 3$

$y = - 4$

$x = - 3$

$y = - 4$

$x = - 3$

$y = - 4$

Step 13

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

$(2, 6)$

$(- 2, - 6)$

$(3, 4)$

$(- 3, - 4)$

Step 14

The result can be shown in multiple forms.

Point Form:

$(2, 6), (- 2, - 6), (3, 4), (- 3, - 4)$

Equation Form:

$x = 2, y = 6$

$x = - 2, y = - 6$

$x = 3, y = 4$

$x = - 3, y = - 4$

Step 15

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