Solve by Substitution xy=2 , x^2+y^2=5

$x y = 2$ , $x^{2} + y^{2} = 5$

Step 1

Divide each term in $x y = 2$ by $y$ and simplify.

Divide each term in $x y = 2$ by $y$ .

$\frac{x y}{y} = \frac{2}{y}$

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

Simplify the left side.

Cancel the common factor of $y$ .

Cancel the common factor.

$\frac{x y}{y} = \frac{2}{y}$

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

Divide $x$ by $1$ .

$x = \frac{2}{y}$

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

$x = \frac{2}{y}$

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

$x = \frac{2}{y}$

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

$x = \frac{2}{y}$

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

Step 2

Replace all occurrences of $x$ with $\frac{2}{y}$ in each equation.

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

${(\frac{2}{y})}^{2} + y^{2} = 5$

$x = \frac{2}{y}$

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{2}{y}$ .

$\frac{2^{2}}{y^{2}} + y^{2} = 5$

$x = \frac{2}{y}$

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

$\frac{4}{y^{2}} + y^{2} = 5$

$x = \frac{2}{y}$

$\frac{4}{y^{2}} + y^{2} = 5$

$x = \frac{2}{y}$

$\frac{4}{y^{2}} + y^{2} = 5$

$x = \frac{2}{y}$

$\frac{4}{y^{2}} + y^{2} = 5$

$x = \frac{2}{y}$

Step 3

Solve for $y$ in $\frac{4}{y^{2}} + y^{2} = 5$ .

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.

$y^{2}, 1, 1$

$x = \frac{2}{y}$

The LCM of one and any expression is the expression.

$y^{2}$

$x = \frac{2}{y}$

$y^{2}$

$x = \frac{2}{y}$

Multiply each term in $\frac{4}{y^{2}} + y^{2} = 5$ by $y^{2}$ to eliminate the fractions.

Multiply each term in $\frac{4}{y^{2}} + y^{2} = 5$ by $y^{2}$ .

$\frac{4}{y^{2}} \cdot y^{2} + y^{2} y^{2} = 5 y^{2}$

$x = \frac{2}{y}$

Simplify the left side.

Simplify each term.

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

Cancel the common factor.

$\frac{4}{y^{2}} \cdot y^{2} + y^{2} y^{2} = 5 y^{2}$

$x = \frac{2}{y}$

Rewrite the expression.

$4 + y^{2} y^{2} = 5 y^{2}$

$x = \frac{2}{y}$

$4 + y^{2} y^{2} = 5 y^{2}$

$x = \frac{2}{y}$

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

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

$4 + y^{2 + 2} = 5 y^{2}$

$x = \frac{2}{y}$

Add $2$ and $2$ .

$4 + y^{4} = 5 y^{2}$

$x = \frac{2}{y}$

$4 + y^{4} = 5 y^{2}$

$x = \frac{2}{y}$

$4 + y^{4} = 5 y^{2}$

$x = \frac{2}{y}$

$4 + y^{4} = 5 y^{2}$

$x = \frac{2}{y}$

$4 + y^{4} = 5 y^{2}$

$x = \frac{2}{y}$

Solve the equation.

Subtract $5 y^{2}$ from both sides of the equation.

$4 + y^{4} - 5 y^{2} = 0$

$x = \frac{2}{y}$

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

$u^{2} - 5 u + 4 = 0$

$u = y^{2}$

$x = \frac{2}{y}$

Factor $u^{2} - 5 u + 4$ 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 $4$ and whose sum is $- 5$ .

$- 4, - 1$

$x = \frac{2}{y}$

Write the factored form using these integers.

$(u - 4) (u - 1) = 0$

$x = \frac{2}{y}$

$(u - 4) (u - 1) = 0$

$x = \frac{2}{y}$

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

$u - 4 = 0$

$u - 1 = 0$

$x = \frac{2}{y}$

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

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

$u - 4 = 0$

$x = \frac{2}{y}$

Add $4$ to both sides of the equation.

$u = 4$

$x = \frac{2}{y}$

$u = 4$

$x = \frac{2}{y}$

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

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

$u - 1 = 0$

$x = \frac{2}{y}$

Add $1$ to both sides of the equation.

$u = 1$

$x = \frac{2}{y}$

$u = 1$

$x = \frac{2}{y}$

The final solution is all the values that make $(u - 4) (u - 1) = 0$ true.

$u = 4, 1$

$x = \frac{2}{y}$

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

$y^{2} = 4$

$y^{2} = 1$

$x = \frac{2}{y}$

Solve the first equation for $y$ .

$y^{2} = 4$

$x = \frac{2}{y}$

Solve the equation for $y$ .

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

$y = \pm \sqrt{4}$

$x = \frac{2}{y}$

Simplify $\pm \sqrt{4}$ .

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

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

$x = \frac{2}{y}$

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

$y = \pm 2$

$x = \frac{2}{y}$

$y = \pm 2$

$x = \frac{2}{y}$

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.

$y = 2$

$x = \frac{2}{y}$

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

$y = - 2$

$x = \frac{2}{y}$

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

$y = 2, - 2$

$x = \frac{2}{y}$

$y = 2, - 2$

$x = \frac{2}{y}$

$y = 2, - 2$

$x = \frac{2}{y}$

Solve the second equation for $y$ .

$y^{2} = 1$

$x = \frac{2}{y}$

Solve the equation for $y$ .

Remove parentheses.

$y^{2} = 1$

$x = \frac{2}{y}$

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

$y = \pm \sqrt{1}$

$x = \frac{2}{y}$

Any root of $1$ is $1$ .

$y = \pm 1$

$x = \frac{2}{y}$

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.

$y = 1$

$x = \frac{2}{y}$

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

$y = - 1$

$x = \frac{2}{y}$

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

$y = 1, - 1$

$x = \frac{2}{y}$

$y = 1, - 1$

$x = \frac{2}{y}$

$y = 1, - 1$

$x = \frac{2}{y}$

The solution to $y^{4} - 5 y^{2} + 4 = 0$ is $y = 2, - 2, 1, - 1$ .

$y = 2, - 2, 1, - 1$

$x = \frac{2}{y}$

$y = 2, - 2, 1, - 1$

$x = \frac{2}{y}$

$y = 2, - 2, 1, - 1$

$x = \frac{2}{y}$

Step 4

Replace all occurrences of $y$ with $2$ in each equation.

Replace all occurrences of $y$ in $x = \frac{2}{y}$ with $2$ .

$x = \frac{2}{2}$

$y = 2$

Simplify the right side.

Divide $2$ by $2$ .

$x = 1$

$y = 2$

$x = 1$

$y = 2$

$x = 1$

$y = 2$

Step 5

Replace all occurrences of $y$ with $- 2$ in each equation.

Replace all occurrences of $y$ in $x = \frac{2}{y}$ with $- 2$ .

$x = \frac{2}{- 2}$

$y = - 2$

Simplify the right side.

Divide $2$ by $- 2$ .

$x = - 1$

$y = - 2$

$x = - 1$

$y = - 2$

$x = - 1$

$y = - 2$

Step 6

Replace all occurrences of $y$ with $2$ in each equation.

Replace all occurrences of $y$ in $x = \frac{2}{y}$ with $2$ .

$x = \frac{2}{2}$

$y = 2$

Simplify the right side.

Divide $2$ by $2$ .

$x = 1$

$y = 2$

$x = 1$

$y = 2$

$x = 1$

$y = 2$

Step 7

Replace all occurrences of $y$ with $- 2$ in each equation.

Replace all occurrences of $y$ in $x = \frac{2}{y}$ with $- 2$ .

$x = \frac{2}{- 2}$

$y = - 2$

Simplify the right side.

Divide $2$ by $- 2$ .

$x = - 1$

$y = - 2$

$x = - 1$

$y = - 2$

$x = - 1$

$y = - 2$

Step 8

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

Replace all occurrences of $y$ in $x = \frac{2}{y}$ with $1$ .

$x = \frac{2}{1}$

$y = 1$

Simplify the right side.

Divide $2$ by $1$ .

$x = 2$

$y = 1$

$x = 2$

$y = 1$

$x = 2$

$y = 1$

Step 9

Replace all occurrences of $y$ with $2$ in each equation.

Replace all occurrences of $y$ in $x = \frac{2}{y}$ with $2$ .

$x = \frac{2}{2}$

$y = 2$

Simplify the right side.

Divide $2$ by $2$ .

$x = 1$

$y = 2$

$x = 1$

$y = 2$

$x = 1$

$y = 2$

Step 10

Replace all occurrences of $y$ with $- 2$ in each equation.

Replace all occurrences of $y$ in $x = \frac{2}{y}$ with $- 2$ .

$x = \frac{2}{- 2}$

$y = - 2$

Simplify the right side.

Divide $2$ by $- 2$ .

$x = - 1$

$y = - 2$

$x = - 1$

$y = - 2$

$x = - 1$

$y = - 2$

Step 11

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

Replace all occurrences of $y$ in $x = \frac{2}{y}$ with $1$ .

$x = \frac{2}{1}$

$y = 1$

Simplify the right side.

Divide $2$ by $1$ .

$x = 2$

$y = 1$

$x = 2$

$y = 1$

$x = 2$

$y = 1$

Step 12

Replace all occurrences of $y$ with $- 1$ in each equation.

Replace all occurrences of $y$ in $x = \frac{2}{y}$ with $- 1$ .

$x = \frac{2}{- 1}$

$y = - 1$

Simplify the right side.

Divide $2$ by $- 1$ .

$x = - 2$

$y = - 1$

$x = - 2$

$y = - 1$

$x = - 2$

$y = - 1$

Step 13

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

$(1, 2)$

$(- 1, - 2)$

$(2, 1)$

$(- 2, - 1)$

Step 14

The result can be shown in multiple forms.

Point Form:

$(1, 2), (- 1, - 2), (2, 1), (- 2, - 1)$

Equation Form:

$x = 1, y = 2$

$x = - 1, y = - 2$

$x = 2, y = 1$

$x = - 2, y = - 1$

Step 15

Solve by Substitution xy=2 , x^2+y^2=5

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