Solved on Sep 25, 2023

Solve the system of nonlinear equations: $x^2 + y^2 = 26$ and $3x^2 + 25y^2 = 100$ to find the four (4) solutions.

STEP 1

Assumptions1. We are given two equations, $x^{}+y^{}=26$ and $3x^{}+25y^{}=100$ . . We are asked to solve this system of nonlinear equations.
3. We are told that there are four solutions.

STEP 2

First, we can express $y^{2}$ from the first equation and substitute it into the second equation. This will give us a single equation in terms of $x$ .
$y^{2} =26 - x^{2}$ Substitute $y^{2}$ into the second equation $x^{2} +25(26 - x^{2}) =100$

STEP 3

implify the equation by distributing the25 $3x^{2} +650 -25x^{2} =100$

STEP 4

Combine like terms $-22x^{2} +650 =100$

STEP 5

Subtract650 from both sides $-22x^{2} = -550$

STEP 6

Divide both sides by -22 $x^{2} =25$

STEP 7

Take the square root of both sides. Remember that the square root of a number can be either positive or negative, so we have two possible values for $x$ $x = \pm5$

STEP 8

Substitute $x =5$ and $x = -5$ into the first equation to find the corresponding $y$ valuesFor $x =5$ $y^{2} =26 -5^{2} =26 -25 =1$ So, $y = \pm1$ .
For $x = -5$ $y^{2} =26 - (-5)^{2} =26 -25 =1$ So, $y = \pm1$ .

STEP 9

Therefore, the four solutions to the system of equations are $(x, y) = (5,), (5, -), (-5,), (-5, -)$

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Solve the system of nonlinear equations: $x^2 + y^2 = 26$ and $3x^2 + 25y^2 = 100$ to find the four (4) solutions.

STEP 1

Assumptions1. We are given two equations, $x^{}+y^{}=26$ and $3x^{}+25y^{}=100$ . . We are asked to solve this system of nonlinear equations.
3. We are told that there are four solutions.

STEP 2

First, we can express $y^{2}$ from the first equation and substitute it into the second equation. This will give us a single equation in terms of $x$ .
$y^{2} =26 - x^{2}$ Substitute $y^{2}$ into the second equation $x^{2} +25(26 - x^{2}) =100$

STEP 3

implify the equation by distributing the25 $3x^{2} +650 -25x^{2} =100$

STEP 4

Combine like terms $-22x^{2} +650 =100$

STEP 5

Subtract650 from both sides $-22x^{2} = -550$

STEP 6

Divide both sides by -22 $x^{2} =25$

STEP 7

Take the square root of both sides. Remember that the square root of a number can be either positive or negative, so we have two possible values for $x$ $x = \pm5$

STEP 8

Substitute $x =5$ and $x = -5$ into the first equation to find the corresponding $y$ valuesFor $x =5$ $y^{2} =26 -5^{2} =26 -25 =1$ So, $y = \pm1$ .
For $x = -5$ $y^{2} =26 - (-5)^{2} =26 -25 =1$ So, $y = \pm1$ .

STEP 9

Therefore, the four solutions to the system of equations are $(x, y) = (5,), (5, -), (-5,), (-5, -)$

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Solve the system of nonlinear equations: x2+y2=26x^2 + y^2 = 26x2+y2=26 and 3x2+25y2=1003x^2 + 25y^2 = 1003x2+25y2=100 to find the four (4) solutions.

STEP 1

Assumptions1. We are given two equations, x+y=26x^{}+y^{}=26x+y=26 and 3x+25y=1003x^{}+25y^{}=1003x+25y=100. . We are asked to solve this system of nonlinear equations.3. We are told that there are four solutions.

STEP 2

First, we can express y2y^{2}y2 from the first equation and substitute it into the second equation. This will give us a single equation in terms of xxx.y2=26−x2y^{2} =26 - x^{2}y2=26−x2Substitute y2y^{2}y2 into the second equationx2+25(26−x2)=100x^{2} +25(26 - x^{2}) =100x2+25(26−x2)=100

STEP 3

implify the equation by distributing the253x2+650−25x2=1003x^{2} +650 -25x^{2} =1003x2+650−25x2=100

STEP 4

Combine like terms−22x2+650=100-22x^{2} +650 =100−22x2+650=100

STEP 5

Subtract650 from both sides−22x2=−550-22x^{2} = -550−22x2=−550

STEP 6

Divide both sides by -22x2=25x^{2} =25x2=25

STEP 7

Take the square root of both sides. Remember that the square root of a number can be either positive or negative, so we have two possible values for xxxx=±5x = \pm5x=±5

STEP 8

STEP 9

Therefore, the four solutions to the system of equations are(x,y)=(5,),(5,−),(−5,),(−5,−)(x, y) = (5,), (5, -), (-5,), (-5, -)(x,y)=(5,),(5,−),(−5,),(−5,−)

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Solve the system of nonlinear equations: $x^2 + y^2 = 26$ and $3x^2 + 25y^2 = 100$ to find the four (4) solutions.

Assumptions1. We are given two equations, $x^{}+y^{}=26$ and $3x^{}+25y^{}=100$ . . We are asked to solve this system of nonlinear equations.
3. We are told that there are four solutions.

First, we can express $y^{2}$ from the first equation and substitute it into the second equation. This will give us a single equation in terms of $x$ .
$y^{2} =26 - x^{2}$ Substitute $y^{2}$ into the second equation $x^{2} +25(26 - x^{2}) =100$

implify the equation by distributing the25 $3x^{2} +650 -25x^{2} =100$

Combine like terms $-22x^{2} +650 =100$

Subtract650 from both sides $-22x^{2} = -550$

Divide both sides by -22 $x^{2} =25$

Take the square root of both sides. Remember that the square root of a number can be either positive or negative, so we have two possible values for $x$ $x = \pm5$

Therefore, the four solutions to the system of equations are $(x, y) = (5,), (5, -), (-5,), (-5, -)$