QuestionSolve the system: $2x^{2}+y^{2}=5$ and $x^{2}-2y^{2}+12=0$ . List all solutions or state if none exist.

Studdy Solution

STEP 1

Assumptions1. We have a system of two equations $x^{}+y^{}=5$ $x^{}- y^{}+120$ . We need to find all the solutions for this system.

STEP 2

To solve this system, we can use the method of substitution or elimination. Here, we will use substitution. First, let's solve the second equation for $x^{2}$ .
$x^{2}=2 y^{2}-12$

STEP 3

Now, substitute $x^{2}$ from the second equation into the first equation.
$2(2 y^{2}-12)+y^{2}=5$

STEP 4

implify the equation.
$4 y^{2}-24+y^{2}=$

STEP 5

Combine like terms.
$5 y^{2}-24=5$

STEP 6

Add24 to both sides of the equation.
$5 y^{2}=29$

STEP 7

Divide both sides by5.
$y^{2}=\frac{29}{5}$

STEP 8

Take the square root of both sides. Remember that the square root of a number can be positive or negative.
$y=\pm\sqrt{\frac{29}{5}}$

STEP 9

Now, substitute $y$ into the second equation to solve for $x$ .
$x^{2}=2 (\pm\sqrt{\frac{29}{5}})^{2}-12$

STEP 10

implify the equation.
$x^{2}=2 \cdot \frac{29}{5}-12$

STEP 11

Multiply the terms.
$x^{}=\frac{58}{5}-$

STEP 12

Convert12 into a fraction with5 as the denominator.
$x^{2}=\frac{58}{5}-\frac{60}{5}$

STEP 13

Subtract the fractions.
$x^{2}=\frac{-2}{5}$

STEP 14

Take the square root of both sides. Remember that the square root of a number can be positive or negative.
$x=\pm\sqrt{\frac{-2}{}}$ However, the square root of a negative number is not a real number, so there are no real solutions for $x$ .
Therefore, the system has no solution.

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QuestionSolve the system: $2x^{2}+y^{2}=5$ and $x^{2}-2y^{2}+12=0$ . List all solutions or state if none exist.

Studdy Solution

STEP 1

Assumptions1. We have a system of two equations $x^{}+y^{}=5$ $x^{}- y^{}+120$ . We need to find all the solutions for this system.

STEP 2

To solve this system, we can use the method of substitution or elimination. Here, we will use substitution. First, let's solve the second equation for $x^{2}$ .
$x^{2}=2 y^{2}-12$

STEP 3

Now, substitute $x^{2}$ from the second equation into the first equation.
$2(2 y^{2}-12)+y^{2}=5$

STEP 4

implify the equation.
$4 y^{2}-24+y^{2}=$

STEP 5

Combine like terms.
$5 y^{2}-24=5$

STEP 6

Add24 to both sides of the equation.
$5 y^{2}=29$

STEP 7

Divide both sides by5.
$y^{2}=\frac{29}{5}$

STEP 8

Take the square root of both sides. Remember that the square root of a number can be positive or negative.
$y=\pm\sqrt{\frac{29}{5}}$

STEP 9

Now, substitute $y$ into the second equation to solve for $x$ .
$x^{2}=2 (\pm\sqrt{\frac{29}{5}})^{2}-12$

STEP 10

implify the equation.
$x^{2}=2 \cdot \frac{29}{5}-12$

STEP 11

Multiply the terms.
$x^{}=\frac{58}{5}-$

STEP 12

Convert12 into a fraction with5 as the denominator.
$x^{2}=\frac{58}{5}-\frac{60}{5}$

STEP 13

Subtract the fractions.
$x^{2}=\frac{-2}{5}$

STEP 14

Take the square root of both sides. Remember that the square root of a number can be positive or negative.
$x=\pm\sqrt{\frac{-2}{}}$ However, the square root of a negative number is not a real number, so there are no real solutions for $x$ .
Therefore, the system has no solution.

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QuestionSolve the system: 2x2+y2=52x^{2}+y^{2}=52x2+y2=5 and x2−2y2+12=0x^{2}-2y^{2}+12=0x2−2y2+12=0. List all solutions or state if none exist.

Studdy Solution

STEP 1

Assumptions1. We have a system of two equations x+y=5 x^{}+y^{}=5x+y=5 x−y+120x^{}- y^{}+120x−y+120. We need to find all the solutions for this system.

STEP 2

To solve this system, we can use the method of substitution or elimination. Here, we will use substitution. First, let's solve the second equation for x2x^{2}x2.x2=2y2−12x^{2}=2 y^{2}-12x2=2y2−12

STEP 3

Now, substitute x2x^{2}x2 from the second equation into the first equation.2(2y2−12)+y2=52(2 y^{2}-12)+y^{2}=52(2y2−12)+y2=5

STEP 4

implify the equation.4y2−24+y2=4 y^{2}-24+y^{2}=4y2−24+y2=

STEP 5

Combine like terms.5y2−24=55 y^{2}-24=55y2−24=5

STEP 6

Add24 to both sides of the equation.5y2=295 y^{2}=295y2=29

STEP 7

Divide both sides by5.y2=295y^{2}=\frac{29}{5}y2=529​

STEP 8

Take the square root of both sides. Remember that the square root of a number can be positive or negative.y=±295y=\pm\sqrt{\frac{29}{5}}y=±529​​

STEP 9

Now, substitute yyy into the second equation to solve for xxx.x2=2(±295)2−12x^{2}=2 (\pm\sqrt{\frac{29}{5}})^{2}-12x2=2(±529​​)2−12

STEP 10

implify the equation.x2=2⋅295−12x^{2}=2 \cdot \frac{29}{5}-12x2=2⋅529​−12

STEP 11

Multiply the terms.x=585−x^{}=\frac{58}{5}-x=558​−

STEP 12

Convert12 into a fraction with5 as the denominator.x2=585−605x^{2}=\frac{58}{5}-\frac{60}{5}x2=558​−560​

STEP 13

Subtract the fractions.x2=−25x^{2}=\frac{-2}{5}x2=5−2​

STEP 14

Take the square root of both sides. Remember that the square root of a number can be positive or negative.x=±−2x=\pm\sqrt{\frac{-2}{}}x=±−2​​However, the square root of a negative number is not a real number, so there are no real solutions for xxx.Therefore, the system has no solution.

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QuestionSolve the system: $2x^{2}+y^{2}=5$ and $x^{2}-2y^{2}+12=0$ . List all solutions or state if none exist.

Assumptions1. We have a system of two equations $x^{}+y^{}=5$ $x^{}- y^{}+120$ . We need to find all the solutions for this system.

To solve this system, we can use the method of substitution or elimination. Here, we will use substitution. First, let's solve the second equation for $x^{2}$ .
$x^{2}=2 y^{2}-12$

Now, substitute $x^{2}$ from the second equation into the first equation.
$2(2 y^{2}-12)+y^{2}=5$

implify the equation.
$4 y^{2}-24+y^{2}=$

Combine like terms.
$5 y^{2}-24=5$

Add24 to both sides of the equation.
$5 y^{2}=29$

Divide both sides by5.
$y^{2}=\frac{29}{5}$

Take the square root of both sides. Remember that the square root of a number can be positive or negative.
$y=\pm\sqrt{\frac{29}{5}}$

Now, substitute $y$ into the second equation to solve for $x$ .
$x^{2}=2 (\pm\sqrt{\frac{29}{5}})^{2}-12$

implify the equation.
$x^{2}=2 \cdot \frac{29}{5}-12$

Multiply the terms.
$x^{}=\frac{58}{5}-$

Convert12 into a fraction with5 as the denominator.
$x^{2}=\frac{58}{5}-\frac{60}{5}$

Subtract the fractions.
$x^{2}=\frac{-2}{5}$

Take the square root of both sides. Remember that the square root of a number can be positive or negative.
$x=\pm\sqrt{\frac{-2}{}}$ However, the square root of a negative number is not a real number, so there are no real solutions for $x$ .
Therefore, the system has no solution.