Non Linear Systems - College Algebra

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Question

Solve the following system of equations

$\1: $x^2$$+y^2$=9\2: 3x+y=1$

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Answer

First lets solve equation 2 for y.

$\3x+y=1\y=1-3x$

Now we plug in into equation 1.

FOIL the left hand side, simplify and then subtract from each side.

Now use the quadratic equation in order to solve for .

Recall the quadratic formula

, where refer to the coefficients in the quadratic equation .

$x=\frac{6\pm\sqrt{36+320}$}{20}$

$x=\frac{6\pm\sqrt{356}$}{20}$

$x=\frac{6\pm\sqrt{4\cdot 89}$}{20}$

$x=\frac{6\pm2\sqrt{89}$}{20}$

So the solutions for x are as follows

$x=\frac{6+2\sqrt{89}$}{20}=\frac{3+\sqrt{89}$}{10}$

$x=\frac{6-2\sqrt{89}$}{20}=\frac{3-\sqrt{89}$}{10}$

Now we just plug these values of x into equation 2 to solve for y.

$y=1-3($\frac{3+\sqrt{89}$}{10})$

$y=\frac{10}{10}$-($\frac{9+3\sqrt{89}$}{10})=\frac{1+3\sqrt{89}$}{10}$

$y=1-3($\frac{3-\sqrt{89}$}{10})$

$y=\frac{10}{10}$-($\frac{9-3\sqrt{89}$}{10})=\frac{1-3\sqrt{89}$}{10}$

So the solutions to these pair of equations are:

$x=\frac{3+\sqrt{89}$}{10}$

$x=\frac{3-\sqrt{89}$}{10}$

$y=\frac{1+3\sqrt{89}$}{10}$

$y=\frac{1-3\sqrt{89}$}{10}$

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Question

Solve the following system of equations:

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Answer

Subtract the second equation from the first, to eliminate the y terms:

-

This yields the following:

Factor out an x from both terms:

Solve for x:

Plug in the values for x into the first equation:

Therefore, when , , and when ,

Plug in the values for x and y to make sure they satisfy the second equation too:

Solution:

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Question

Solve the following system of equations:

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Answer

Substitute the value of x from the second equation, into the first equation:

Subtract y from both sides of the equation:

Use the quadratic formula to solve for y: $y=\frac{-b\pm $\sqrt{b^{2}$$-4ac}}{2a}=\frac{1\pm \sqrt{1-4(1)(-1)}$}{2}=\frac{1\pm \sqrt{5}$}{2}$

$y=\frac{1+\sqrt{5}$}{2}$ or $$\frac{1-\sqrt{5}$}{2}$

Plug the first value of y into the first equation: $$\frac{1+\sqrt{5}$}{2}=x-1$

Re-write the equation to make it easier to solve for x: $\frac{1+\sqrt{5}$}{2}=x-$\frac{2}{2}$

Add $\frac{2}{2}$ to each side of the equation: $x=\frac{3+\sqrt{5}$}{2}$

The second value of y into the first equation: $$\frac{1-\sqrt{5}$}{2}=x-1$

Re-write the equation to make it easier to solve for x: $\frac{1-\sqrt{5}$}{2}=x-$\frac{2}{2}$

Add $\frac{2}{2}$ to each side of the equation: $x=\frac{3-\sqrt{5}$}{2}$ .

Plug the values of x and y into the second equation to make sure they satisfy both of them:

Solution: $($\frac{3+\sqrt{5}$}{2},$\frac{1+\sqrt{5}$}{2})($\frac{3-\sqrt{5}$}{2},$\frac{1-\sqrt{5}$}{2})$

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Question

Solve the following system of equations:

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Answer

Multiply the second equation by 6, to eliminate the denominator:

Simplify:

Plug the value of y, from the first equation, into the second equation:

Simplify:

Simplify further:

Use the quadratic formula to solve for x: $x=\frac{6\pm \sqrt{36-4(5)(-3)}$}{10}=\frac{6\pm \sqrt{96}$}{10}=\frac{6\pm 4\sqrt{6}$}{10}=\frac{3\pm 2\sqrt{6}$}{5}$

$x=\frac{3+2\sqrt{6}$}{5}$ or $x=\frac{3-2\sqrt{6}$}{5}$

Plug the first value of x into the first equation: $y=\frac{3+2\sqrt{6}$}{5}-1=\frac{3+2\sqrt{6}$}{5}-$\frac{5}{5}$=\frac{2\sqrt{6}$-2}{5}$

Plug the second value of x into the first equation: $y=\frac{3-2\sqrt{6}$}{5}-1=\frac{3-2\sqrt{6}$}{5}-$\frac{5}{5}$=\frac{-2\sqrt{6}$-2}{5}$

Plug the values of x and y into the second equation to make sure they satisfy both of them:

$$\frac{11+4\sqrt{6}$}{25}+\frac{14-4\sqrt{6}$}{25}=\frac{25}{25}$=1$

Solution: $($\frac{3+2\sqrt{6}$}{5},$\frac{2\sqrt{6}$-2}{5})($\frac{3-2\sqrt{6}$}{5},$\frac{-2\sqrt{6}$-2}{5})$

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Question

Solve the following system of equations:

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Answer

Substitute the value of y from the second equation, into the first equation:

Subtract x from both sides of the equation:

Use the quadratic formula to solve for x: $x=\frac{5\pm \sqrt{25-4(2)(2)}$}{4}=\frac{5\pm \sqrt{25-16}$}{4}=\frac{5\pm \sqrt{9}$}{4}=\frac{5\pm 3}{4}$$

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

Plug the first value of x into the second equation:

Plug the second value of x into the second equation: $y=-($\frac{1}{2}$)+1=\frac{1}{2}$$

or $x=\frac{1}{2}$,y=\frac{1}{2}$$

Verify the first solution by plugging the first values of x and y into both equations:

Verify the second solution by plugging the second values of x and y into both equations:

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

Solution:

$(2,-1)($\frac{1}{2}$,$\frac{1}{2}$)$

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Question

Solve the system of equations to find the points where the graphs of the equations intersect.

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Answer

Solve the system of equations,

(1)

(2)

Notice that we can multiply both sides of equation (2) by to obtain,

(3)

Now equate equations (3) and (4) to obtain,

This is a quadratic equation, rearrange to find the roots by setting to zero and factoring,

$x = \left \{ -1, 2\right \}$

Use these values to find the corresponding values of by substituting into either of the original equations. Using equation (1),

$y_1 = $\frac{1}{4}$(-1+3)=\frac{1}{2}$$

$y_2 =\frac{1}{4}$(2+3)=\frac{5}{4}$$

$y = \left \{ $\frac{1}{2}$, $\frac{5}{4}\right \}$

The points $\left(-1, $\frac{1}{2}\right)$ and $\left(2, $\frac{5}{4}$ \right)$ are the intersection points of the two functions,

$y=\frac{1}{4}$(x+3)$

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Question

Solve the following system of equations

$\1: $x^2$$+y^2$=9\2: 3x+y=1$

Tap to reveal answer

Answer

First lets solve equation 2 for y.

$\3x+y=1\y=1-3x$

Now we plug in into equation 1.

FOIL the left hand side, simplify and then subtract from each side.

Now use the quadratic equation in order to solve for .

Recall the quadratic formula

, where refer to the coefficients in the quadratic equation .

$x=\frac{6\pm\sqrt{36+320}$}{20}$

$x=\frac{6\pm\sqrt{356}$}{20}$

$x=\frac{6\pm\sqrt{4\cdot 89}$}{20}$

$x=\frac{6\pm2\sqrt{89}$}{20}$

So the solutions for x are as follows

$x=\frac{6+2\sqrt{89}$}{20}=\frac{3+\sqrt{89}$}{10}$

$x=\frac{6-2\sqrt{89}$}{20}=\frac{3-\sqrt{89}$}{10}$

Now we just plug these values of x into equation 2 to solve for y.

$y=1-3($\frac{3+\sqrt{89}$}{10})$

$y=\frac{10}{10}$-($\frac{9+3\sqrt{89}$}{10})=\frac{1+3\sqrt{89}$}{10}$

$y=1-3($\frac{3-\sqrt{89}$}{10})$

$y=\frac{10}{10}$-($\frac{9-3\sqrt{89}$}{10})=\frac{1-3\sqrt{89}$}{10}$

So the solutions to these pair of equations are:

$x=\frac{3+\sqrt{89}$}{10}$

$x=\frac{3-\sqrt{89}$}{10}$

$y=\frac{1+3\sqrt{89}$}{10}$

$y=\frac{1-3\sqrt{89}$}{10}$

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Question

Solve the following system of equations:

Tap to reveal answer

Answer

Subtract the second equation from the first, to eliminate the y terms:

-

This yields the following:

Factor out an x from both terms:

Solve for x:

Plug in the values for x into the first equation:

Therefore, when , , and when ,

Plug in the values for x and y to make sure they satisfy the second equation too:

Solution:

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Question

Solve the following system of equations:

Tap to reveal answer

Answer

Substitute the value of x from the second equation, into the first equation:

Subtract y from both sides of the equation:

Use the quadratic formula to solve for y: $y=\frac{-b\pm $\sqrt{b^{2}$$-4ac}}{2a}=\frac{1\pm \sqrt{1-4(1)(-1)}$}{2}=\frac{1\pm \sqrt{5}$}{2}$

$y=\frac{1+\sqrt{5}$}{2}$ or $$\frac{1-\sqrt{5}$}{2}$

Plug the first value of y into the first equation: $$\frac{1+\sqrt{5}$}{2}=x-1$

Re-write the equation to make it easier to solve for x: $\frac{1+\sqrt{5}$}{2}=x-$\frac{2}{2}$

Add $\frac{2}{2}$ to each side of the equation: $x=\frac{3+\sqrt{5}$}{2}$

The second value of y into the first equation: $$\frac{1-\sqrt{5}$}{2}=x-1$

Re-write the equation to make it easier to solve for x: $\frac{1-\sqrt{5}$}{2}=x-$\frac{2}{2}$

Add $\frac{2}{2}$ to each side of the equation: $x=\frac{3-\sqrt{5}$}{2}$ .

Plug the values of x and y into the second equation to make sure they satisfy both of them:

Solution: $($\frac{3+\sqrt{5}$}{2},$\frac{1+\sqrt{5}$}{2})($\frac{3-\sqrt{5}$}{2},$\frac{1-\sqrt{5}$}{2})$

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Question

Solve the following system of equations:

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Answer

Multiply the second equation by 6, to eliminate the denominator:

Simplify:

Plug the value of y, from the first equation, into the second equation:

Simplify:

Simplify further:

Use the quadratic formula to solve for x: $x=\frac{6\pm \sqrt{36-4(5)(-3)}$}{10}=\frac{6\pm \sqrt{96}$}{10}=\frac{6\pm 4\sqrt{6}$}{10}=\frac{3\pm 2\sqrt{6}$}{5}$

$x=\frac{3+2\sqrt{6}$}{5}$ or $x=\frac{3-2\sqrt{6}$}{5}$

Plug the first value of x into the first equation: $y=\frac{3+2\sqrt{6}$}{5}-1=\frac{3+2\sqrt{6}$}{5}-$\frac{5}{5}$=\frac{2\sqrt{6}$-2}{5}$

Plug the second value of x into the first equation: $y=\frac{3-2\sqrt{6}$}{5}-1=\frac{3-2\sqrt{6}$}{5}-$\frac{5}{5}$=\frac{-2\sqrt{6}$-2}{5}$

Plug the values of x and y into the second equation to make sure they satisfy both of them:

$$\frac{11+4\sqrt{6}$}{25}+\frac{14-4\sqrt{6}$}{25}=\frac{25}{25}$=1$

Solution: $($\frac{3+2\sqrt{6}$}{5},$\frac{2\sqrt{6}$-2}{5})($\frac{3-2\sqrt{6}$}{5},$\frac{-2\sqrt{6}$-2}{5})$

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Question

Solve the following system of equations:

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Answer

Substitute the value of y from the second equation, into the first equation:

Subtract x from both sides of the equation:

Use the quadratic formula to solve for x: $x=\frac{5\pm \sqrt{25-4(2)(2)}$}{4}=\frac{5\pm \sqrt{25-16}$}{4}=\frac{5\pm \sqrt{9}$}{4}=\frac{5\pm 3}{4}$$

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

Plug the first value of x into the second equation:

Plug the second value of x into the second equation: $y=-($\frac{1}{2}$)+1=\frac{1}{2}$$

or $x=\frac{1}{2}$,y=\frac{1}{2}$$

Verify the first solution by plugging the first values of x and y into both equations:

Verify the second solution by plugging the second values of x and y into both equations:

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

Solution:

$(2,-1)($\frac{1}{2}$,$\frac{1}{2}$)$

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Question

Solve the system of equations to find the points where the graphs of the equations intersect.

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Answer

Solve the system of equations,

(1)

(2)

Notice that we can multiply both sides of equation (2) by to obtain,

(3)

Now equate equations (3) and (4) to obtain,

This is a quadratic equation, rearrange to find the roots by setting to zero and factoring,

$x = \left \{ -1, 2\right \}$

Use these values to find the corresponding values of by substituting into either of the original equations. Using equation (1),

$y_1 = $\frac{1}{4}$(-1+3)=\frac{1}{2}$$

$y_2 =\frac{1}{4}$(2+3)=\frac{5}{4}$$

$y = \left \{ $\frac{1}{2}$, $\frac{5}{4}\right \}$

The points $\left(-1, $\frac{1}{2}\right)$ and $\left(2, $\frac{5}{4}$ \right)$ are the intersection points of the two functions,

$y=\frac{1}{4}$(x+3)$

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Question

Solve the following system of equations

$\1: $x^2$$+y^2$=9\2: 3x+y=1$

Tap to reveal answer

Answer

First lets solve equation 2 for y.

$\3x+y=1\y=1-3x$

Now we plug in into equation 1.

FOIL the left hand side, simplify and then subtract from each side.

Now use the quadratic equation in order to solve for .

Recall the quadratic formula

, where refer to the coefficients in the quadratic equation .

$x=\frac{6\pm\sqrt{36+320}$}{20}$

$x=\frac{6\pm\sqrt{356}$}{20}$

$x=\frac{6\pm\sqrt{4\cdot 89}$}{20}$

$x=\frac{6\pm2\sqrt{89}$}{20}$

So the solutions for x are as follows

$x=\frac{6+2\sqrt{89}$}{20}=\frac{3+\sqrt{89}$}{10}$

$x=\frac{6-2\sqrt{89}$}{20}=\frac{3-\sqrt{89}$}{10}$

Now we just plug these values of x into equation 2 to solve for y.

$y=1-3($\frac{3+\sqrt{89}$}{10})$

$y=\frac{10}{10}$-($\frac{9+3\sqrt{89}$}{10})=\frac{1+3\sqrt{89}$}{10}$

$y=1-3($\frac{3-\sqrt{89}$}{10})$

$y=\frac{10}{10}$-($\frac{9-3\sqrt{89}$}{10})=\frac{1-3\sqrt{89}$}{10}$

So the solutions to these pair of equations are:

$x=\frac{3+\sqrt{89}$}{10}$

$x=\frac{3-\sqrt{89}$}{10}$

$y=\frac{1+3\sqrt{89}$}{10}$

$y=\frac{1-3\sqrt{89}$}{10}$

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Question

Solve the following system of equations:

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Answer

Subtract the second equation from the first, to eliminate the y terms:

-

This yields the following:

Factor out an x from both terms:

Solve for x:

Plug in the values for x into the first equation:

Therefore, when , , and when ,

Plug in the values for x and y to make sure they satisfy the second equation too:

Solution:

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Question

Solve the following system of equations:

Tap to reveal answer

Answer

Substitute the value of x from the second equation, into the first equation:

Subtract y from both sides of the equation:

Use the quadratic formula to solve for y: $y=\frac{-b\pm $\sqrt{b^{2}$$-4ac}}{2a}=\frac{1\pm \sqrt{1-4(1)(-1)}$}{2}=\frac{1\pm \sqrt{5}$}{2}$

$y=\frac{1+\sqrt{5}$}{2}$ or $$\frac{1-\sqrt{5}$}{2}$

Plug the first value of y into the first equation: $$\frac{1+\sqrt{5}$}{2}=x-1$

Re-write the equation to make it easier to solve for x: $\frac{1+\sqrt{5}$}{2}=x-$\frac{2}{2}$

Add $\frac{2}{2}$ to each side of the equation: $x=\frac{3+\sqrt{5}$}{2}$

The second value of y into the first equation: $$\frac{1-\sqrt{5}$}{2}=x-1$

Re-write the equation to make it easier to solve for x: $\frac{1-\sqrt{5}$}{2}=x-$\frac{2}{2}$

Add $\frac{2}{2}$ to each side of the equation: $x=\frac{3-\sqrt{5}$}{2}$ .

Plug the values of x and y into the second equation to make sure they satisfy both of them:

Solution: $($\frac{3+\sqrt{5}$}{2},$\frac{1+\sqrt{5}$}{2})($\frac{3-\sqrt{5}$}{2},$\frac{1-\sqrt{5}$}{2})$

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Question

Solve the following system of equations:

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Answer

Multiply the second equation by 6, to eliminate the denominator:

Simplify:

Plug the value of y, from the first equation, into the second equation:

Simplify:

Simplify further:

Use the quadratic formula to solve for x: $x=\frac{6\pm \sqrt{36-4(5)(-3)}$}{10}=\frac{6\pm \sqrt{96}$}{10}=\frac{6\pm 4\sqrt{6}$}{10}=\frac{3\pm 2\sqrt{6}$}{5}$

$x=\frac{3+2\sqrt{6}$}{5}$ or $x=\frac{3-2\sqrt{6}$}{5}$

Plug the first value of x into the first equation: $y=\frac{3+2\sqrt{6}$}{5}-1=\frac{3+2\sqrt{6}$}{5}-$\frac{5}{5}$=\frac{2\sqrt{6}$-2}{5}$

Plug the second value of x into the first equation: $y=\frac{3-2\sqrt{6}$}{5}-1=\frac{3-2\sqrt{6}$}{5}-$\frac{5}{5}$=\frac{-2\sqrt{6}$-2}{5}$

Plug the values of x and y into the second equation to make sure they satisfy both of them:

$$\frac{11+4\sqrt{6}$}{25}+\frac{14-4\sqrt{6}$}{25}=\frac{25}{25}$=1$

Solution: $($\frac{3+2\sqrt{6}$}{5},$\frac{2\sqrt{6}$-2}{5})($\frac{3-2\sqrt{6}$}{5},$\frac{-2\sqrt{6}$-2}{5})$

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Question

Solve the following system of equations:

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Answer

Substitute the value of y from the second equation, into the first equation:

Subtract x from both sides of the equation:

Use the quadratic formula to solve for x: $x=\frac{5\pm \sqrt{25-4(2)(2)}$}{4}=\frac{5\pm \sqrt{25-16}$}{4}=\frac{5\pm \sqrt{9}$}{4}=\frac{5\pm 3}{4}$$

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

Plug the first value of x into the second equation:

Plug the second value of x into the second equation: $y=-($\frac{1}{2}$)+1=\frac{1}{2}$$

or $x=\frac{1}{2}$,y=\frac{1}{2}$$

Verify the first solution by plugging the first values of x and y into both equations:

Verify the second solution by plugging the second values of x and y into both equations:

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

Solution:

$(2,-1)($\frac{1}{2}$,$\frac{1}{2}$)$

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Question

Solve the system of equations to find the points where the graphs of the equations intersect.

Tap to reveal answer

Answer

Solve the system of equations,

(1)

(2)

Notice that we can multiply both sides of equation (2) by to obtain,

(3)

Now equate equations (3) and (4) to obtain,

This is a quadratic equation, rearrange to find the roots by setting to zero and factoring,

$x = \left \{ -1, 2\right \}$

Use these values to find the corresponding values of by substituting into either of the original equations. Using equation (1),

$y_1 = $\frac{1}{4}$(-1+3)=\frac{1}{2}$$

$y_2 =\frac{1}{4}$(2+3)=\frac{5}{4}$$

$y = \left \{ $\frac{1}{2}$, $\frac{5}{4}\right \}$

The points $\left(-1, $\frac{1}{2}\right)$ and $\left(2, $\frac{5}{4}$ \right)$ are the intersection points of the two functions,

$y=\frac{1}{4}$(x+3)$

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Question

Solve the following system of equations

$\1: $x^2$$+y^2$=9\2: 3x+y=1$

Tap to reveal answer

Answer

First lets solve equation 2 for y.

$\3x+y=1\y=1-3x$

Now we plug in into equation 1.

FOIL the left hand side, simplify and then subtract from each side.

Now use the quadratic equation in order to solve for .

Recall the quadratic formula

, where refer to the coefficients in the quadratic equation .

$x=\frac{6\pm\sqrt{36+320}$}{20}$

$x=\frac{6\pm\sqrt{356}$}{20}$

$x=\frac{6\pm\sqrt{4\cdot 89}$}{20}$

$x=\frac{6\pm2\sqrt{89}$}{20}$

So the solutions for x are as follows

$x=\frac{6+2\sqrt{89}$}{20}=\frac{3+\sqrt{89}$}{10}$

$x=\frac{6-2\sqrt{89}$}{20}=\frac{3-\sqrt{89}$}{10}$

Now we just plug these values of x into equation 2 to solve for y.

$y=1-3($\frac{3+\sqrt{89}$}{10})$

$y=\frac{10}{10}$-($\frac{9+3\sqrt{89}$}{10})=\frac{1+3\sqrt{89}$}{10}$

$y=1-3($\frac{3-\sqrt{89}$}{10})$

$y=\frac{10}{10}$-($\frac{9-3\sqrt{89}$}{10})=\frac{1-3\sqrt{89}$}{10}$

So the solutions to these pair of equations are:

$x=\frac{3+\sqrt{89}$}{10}$

$x=\frac{3-\sqrt{89}$}{10}$

$y=\frac{1+3\sqrt{89}$}{10}$

$y=\frac{1-3\sqrt{89}$}{10}$

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Question

Solve the following system of equations:

Tap to reveal answer

Answer

Subtract the second equation from the first, to eliminate the y terms:

-

This yields the following:

Factor out an x from both terms:

Solve for x:

Plug in the values for x into the first equation:

Therefore, when , , and when ,

Plug in the values for x and y to make sure they satisfy the second equation too:

Solution:

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