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College Algebra : Systems of Equations

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Example Questions

Example Question #2 : Augmented Matrices

When solving a system of three linear equations in three variables using the Gauss-Jordan elimination method, your initial augmented matrix is as follows:

$\begin{bmatrix}-3 & 2& 3 \\2 & 4 & -3 \\-5 & 2 & 4 \\ \end{matrix}$ $\begin{vmatrix}5 \\0 \\2 \\ \end{bmatrix}$

Which of the following is notation for the next step you should perform?

Possible Answers:

$\frac{1}{4}R3 \rightarrow R3$

$\frac{1}{2}R3 \rightarrow R3$

$\frac{1}{5}R1 \rightarrow R1$

$-\frac{1}{3}R1 \rightarrow R1$

$\frac{1}{4}R2 \rightarrow R2$

Correct answer:

$-\frac{1}{3}R1 \rightarrow R1$

Explanation:

When applying the Gauss-Jordan method to solve a three-by-three linear system, the objective is to use row operations to form an augmented matrix of the form

$\begin{bmatrix}1 & 0 &0 \\ 0 & 1&0 \\0 &0& 1 \end{matrix}$ $\begin{vmatrix}A\\B \\ C \end{bmatrix}$ ,

with (A, B,C) the solution of the system.

Once the initial augmented matrix

$\begin{bmatrix}-3 & 2& 3 \\2 & 4 & -3 \\-5 & 2 & 4 \\ \end{matrix}$ $\begin{vmatrix}5 \\0 \\2 \\ \end{bmatrix}$

is set up, the first step should always be to get a 1 in the upper left position; this is usually done by multiplying every element in Row 1 by the reciprocal of the first element in that row. Since the element in that position is , every element in Row 1 is multiplied by reciprocal $- \frac{1}{3}$ . This step can be written using the notation $-\frac{1}{3}R1 \rightarrow R1$ .

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Example Question #7 : Augmented Matrices

Use augmented matrices to solve the following system of equations:

3x+2y=7

x-y=4

Possible Answers:

$\begin{bmatrix}1&0&\mid &3 & \\ 0 &1&\mid &1 & \end{bmatrix}$

$\begin{bmatrix}1&0&\mid &3 & \\ 0 &1&\mid &-1 & \end{bmatrix}$

$\begin{bmatrix}1&0&\mid &-3 & \\ 0 &1&\mid &1 & \end{bmatrix}$

$\begin{bmatrix}1&0&\mid &-3 & \\ 0 &1&\mid &-1 & \end{bmatrix}$

Correct answer:

$\begin{bmatrix}1&0&\mid &3 & \\ 0 &1&\mid &-1 & \end{bmatrix}$

Explanation:

Begin by constructing an augmented matrix for our system of equations:

$\begin{bmatrix}3&2&\mid &7 & \\ 1 &-1&\mid &4 & \end{bmatrix}$

Switch rows 1 and 2:

$\begin{bmatrix}3&2&\mid &7 & \\ 1 &-1&\mid &4 & \end{bmatrix}$ $R_{1}\Leftrightarrow R_{2}$ $\begin{bmatrix}1&-1&\mid &4 & \\ 3 &2&\mid &7 & \end{bmatrix}$

Add row 2 and twice row 1, to row 1:

$\begin{bmatrix}1&-1&\mid &4 & \\ 3 &2&\mid &7 & \end{bmatrix}$ $2R_{1} \Leftrightarrow R_{2}$ $\begin{bmatrix}5&0&\mid &15 & \\ 3 &2&\mid &7 & \end{bmatrix}$

Divide row 1 by 5:

$\begin{bmatrix}5&0&\mid &15 & \\ 3 &2&\mid &7 & \end{bmatrix}$ $\frac{1}{5}R_{1} \Rightarrow$ $\begin{bmatrix}1&0&\mid &3 & \\ 3 &2&\mid &7 & \end{bmatrix}$

Subtract 3 times row 1, from row 2:

$\begin{bmatrix}1&0&\mid &3 & \\ 3 &2&\mid &7 & \end{bmatrix}$ $R_{2} - 3R_{1} \Rightarrow R_{2}$ $\begin{bmatrix}1&0&\mid &3 & \\ 0 &2&\mid &-2 & \end{bmatrix}$

Divide row 2 by 2:

$\begin{bmatrix}1&0&\mid &3 & \\ 0 &2&\mid &-2 & \end{bmatrix}$ $\frac{1}{2} R_{2} \Rightarrow$ $\begin{bmatrix}1&0&\mid &3 & \\ 0 &1&\mid &-1 & \end{bmatrix}$

Solution:

$\begin{bmatrix}1&0&\mid &3 & \\ 0 &1&\mid &-1 & \end{bmatrix}$

Plug the respective values of x and y into both equations to verify the solution:

3(3)+2(-1)=9-2=7

(3)-(-1)=3+1=4

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Example Question #7 : Augmented Matrices

Use augmented matrices to solve the following system of equations:

x-y+2z=2

2x+3y-z=-1

x-z=-3

Possible Answers:

$\begin{bmatrix}1&0&0&\mid &-1 & \\ 0&1&0&\mid &-1 & \\ 0&0&1&\mid &2 & \end{bmatrix}$

$\begin{bmatrix}1&0&0&\mid &-1 & \\ 0&1&0&\mid &1 & \\ 0&0&1&\mid &2 & \end{bmatrix}$

$\begin{bmatrix}1&0&0&\mid &-1 & \\ 0&1&0&\mid &-1 & \\ 0&0&1&\mid &-2 & \end{bmatrix}$

$\begin{bmatrix}1&0&0&\mid &1 & \\ 0&1&0&\mid &1 & \\ 0&0&1&\mid &2 & \end{bmatrix}$

Correct answer:

$\begin{bmatrix}1&0&0&\mid &-1 & \\ 0&1&0&\mid &1 & \\ 0&0&1&\mid &2 & \end{bmatrix}$

Explanation:

Construct an augmented matrix for our system of equations:

$\begin{bmatrix}1&-1&2&\mid &2 & \\ 2&3&-1&\mid &-1 & \\ 1&0&-1&\mid &-3 & \end{bmatrix}$

Swap rows 1 and 3:

$\begin{bmatrix}1&-1&2&\mid &2 & \\ 2&3&-1&\mid &-1 & \\ 1&0&-1&\mid &-3 & \end{bmatrix}$ $R_{1} \Leftrightarrow R_{3}$ $\begin{bmatrix}1&0&-1&\mid &-3 & \\ 2&3&-1&\mid &-1 & \\ 1&-1&2&\mid &2 & \end{bmatrix}$

Subtract 2 times row 1 from row 2:

$\begin{bmatrix}1&0&-1&\mid &-3 & \\ 2&3&-1&\mid &-1 & \\ 1&-1&2&\mid &2 & \end{bmatrix}$ $R_{2} - 2R_{1} \Rightarrow R_{2}$

$\begin{bmatrix}1&0&-1&\mid &-3 & \\ 0&3&1&\mid &5 & \\ 1&-1&2&\mid &2 & \end{bmatrix}$

Subtract row 1 from row 3:

$\begin{bmatrix}1&0&-1&\mid &-3 & \\ 0&3&1&\mid &5 & \\ 1&-1&2&\mid &2 & \end{bmatrix}$ $R_{3} - R_{1}\Rightarrow R_{3}$

$\begin{bmatrix}1&0&-1&\mid &-3 & \\ 0&3&1&\mid &5 & \\ 0&-1&3&\mid &5 & \end{bmatrix}$

Add row 2 and three times row 3, to row 3:

$\begin{bmatrix}1&0&-1&\mid &-3 & \\ 0&3&1&\mid &5 & \\ 0&-1&3&\mid &5 & \end{bmatrix}$ $R_{2}+3R_{3}\Rightarrow R_{3}$

$\begin{bmatrix}1&0&-1&\mid &-3 & \\ 0&3&1&\mid &5 & \\ 0&0&10&\mid &20 & \end{bmatrix}$

Divide row 3 by 10:

$\begin{bmatrix}1&0&-1&\mid &-3 & \\ 0&3&1&\mid &5 & \\ 0&0&10&\mid &20 & \end{bmatrix}$ $\frac{1}{10} R_{3}\Rightarrow$ $\begin{bmatrix}1&0&-1&\mid &-3 & \\ 0&3&1&\mid &5 & \\ 0&0&1&\mid &2 & \end{bmatrix}$

Add row 3 to row 1:

$\begin{bmatrix}1&0&-1&\mid &-3 & \\ 0&3&1&\mid &5 & \\ 0&0&1&\mid &2 & \end{bmatrix}$ $R_{3}+R_{1}\Rightarrow R_{1}$ $\begin{bmatrix}1&0&0&\mid &-1 & \\ 0&3&1&\mid &5 & \\ 0&0&1&\mid &2 & \end{bmatrix}$

Subtract row 3 from row 2:

$\begin{bmatrix}1&0&0&\mid &-1 & \\ 0&3&1&\mid &5 & \\ 0&0&1&\mid &2 & \end{bmatrix}$ $R_{2}-R_{3}\Rightarrow R_{2}$ $\begin{bmatrix}1&0&0&\mid &-1 & \\ 0&3&0&\mid &3 & \\ 0&0&1&\mid &2 & \end{bmatrix}$

Divide row 2 by 3:

$\begin{bmatrix}1&0&0&\mid &-1 & \\ 0&3&0&\mid &3 & \\ 0&0&1&\mid &2 & \end{bmatrix}$ $\frac{1}{3}R_{2}\Rightarrow$ $\begin{bmatrix}1&0&0&\mid &-1 & \\ 0&1&0&\mid &1 & \\ 0&0&1&\mid &2 & \end{bmatrix}$

Solution:

$\begin{bmatrix}1&0&0&\mid &-1 & \\ 0&1&0&\mid &1 & \\ 0&0&1&\mid &2 & \end{bmatrix}$

Plug the respective values of x, y, and z into all equations to verify the solution:

(-1)-(1)+2(2)=-2+4=2

2(-1)+3(1)-(2)=-2+3-2=1-2=-1

(-1)-(2)=-3

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Example Question #1 : Non Linear Systems

Solve the following system of equations

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

Possible Answers:

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

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

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

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

There are no Real Solutions

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

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

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

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

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

Correct answer:

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

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

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

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

Explanation:

First lets solve equation 2 for y.

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

Now we plug in y=1-3x into equation 1.

x^2+(1-3x)^2=9

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

$\\x^2+(1-3x)^2=9\\x^2+1-6x+9x^2=9\\10x^2-6x-8=0$

Now use the quadratic equation in order to solve for .

Recall the quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ , where a,b,c refer to the coefficients in the quadratic equation ax^2+bx+c=0 .

$x=\frac{6\pm\sqrt{(-6)^2-4(10)(-8)}}{2(10)}$

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

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

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

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

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=1-3x

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

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

y=1-3x

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

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

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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Example Question #2 : Non Linear Systems

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

x+3=4y

$\frac{1}{4}(x^2+1)=y$

Possible Answers:

$\left(\frac{3}{7}, 1\right)$ , $\left( 2, \frac{3}{4} \right )$

$\left(\frac{3}{2}, -1\right)$ , $\left( 2, \frac{11}{2} \right )$

$\left(-1, 2\right)$ , $\left(\frac{1}{2}, \frac{5}{4} \right )$

$\left(2, 1\right)$ , $\left( 1, \frac{3}{4} \right )$

$\left(-1, \frac{1}{2}\right)$ , $\left(2, \frac{5}{4} \right )$

Correct answer:

$\left(-1, \frac{1}{2}\right)$ , $\left(2, \frac{5}{4} \right )$

Explanation:

Solve the system of equations,

x+3=4y (1)

$\frac{1}{4}(x^2+1)=y$ (2)

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

x^2 +1 = 4y (3)

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

x+3 = x^2 +1

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

x^2-x-2=0

(x+1)(x-2)=0

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

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,

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

$y=\frac{1}{4}(x^2+1)$

Intersection

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Example Question #3 : Non Linear Systems

Solve the following system of equations:

$y=x^{2}+1$

y=2x+1

Possible Answers:

(0,1)(5,2)

(0,1)(2,5)

(1,5)(2,1)

(0,5)(1,2)

Correct answer:

(0,1)(2,5)

Explanation:

$y=x^{2}+1$

y=2x+1

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

$y=x^{2}+1$

- y=2x+1

This yields the following: $0=x^{2}-2x$

Factor out an x from both terms: 0=x(x-2)

Solve for x: x=0,2

Plug in the values for x into the first equation:

$y=(0)^{2}+1=1$

$y=(2)^{2}+1=4+1=5$

Therefore, when x=0 , y=1 , and when x=2 , y=5

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

1=2(0)+1=0+1=1

5=2(2)+1=4+1=5

Solution: (0,1)(2,5)

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Example Question #2 : Non Linear Systems

Solve the following system of equations:

y=x-1

$x=y^{2}$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

y=x-1 $x=y^{2}$

Substitute the value of x from the second equation, into the first equation: $y=(y^{2})-1$

Subtract y from both sides of the equation: $0=y^{2}-1-y$

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:

$(\frac{1+\sqrt{5}}{2})^{2}=\frac{1+2\sqrt{5}+5}{4}=\frac{6+2\sqrt{5}}{4}=\frac{3+\sqrt{5}}{2}$

$(\frac{1-\sqrt{5}}{2})^{2}=\frac{1-2\sqrt{5}}{4}=\frac{1-2\sqrt{5}+5}{4}=\frac{6-2\sqrt{5}}{4}=\frac{3-\sqrt{5}}{2}$

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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Example Question #3 : Non Linear Systems

Solve the following system of equations:

y=x-1

$\frac{x^{2}}{3}+\frac{y^{2}}{2}=1$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

y=x-1 $\frac{x^{2}}{3}+\frac{y^{2}}{2}=1$

Multiply the second equation by 6, to eliminate the denominator: $6(\frac{x^{2}}{3}+\frac{y^{2}}{2}=1)6$

Simplify: $2x^{2}+3y^{2}=6$

Plug the value of y, from the first equation, into the second equation: $2x^{2}+3(x-1)^{2}=6$

Simplify: $2x^{2}+3x^{2}-6x+3=6$

Simplify further: $5x^{2}-6x-3=0$

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:

$\frac{(\frac{3+2\sqrt{6}}{5})^{2}}{3}+\frac{(\frac{2\sqrt{6}-2}{5})^{2}}{2}=(\frac{9+12\sqrt{6}+24}{25})(\frac{1}{3})+(\frac{24-8\sqrt{6}+4}{5})(\frac{1}{2})=$

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

$\frac{(\frac{3-2\sqrt{6}}{5})^{2}}{3}+\frac{(\frac{-2\sqrt{6}-2}{5})^{2}}{2}=(\frac{9-12\sqrt{6}+24}{25})(\frac{1}{3})+(\frac{24+8\sqrt{6}+4}{25})(\frac{1}{2})=\frac{11-4\sqrt{6}}{25}+\frac{14+4\sqrt{6}}{25}=\frac{25}{25}=1$

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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Example Question #2 : Non Linear Systems

Solve the following system of equations:

$x=2y^{2}$

y=-x+1

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

$x=2y^{2}$

y=-x+1

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

$x=2y^{2}=2(-x+1)^{2}=2(x^{2}-2x+1)=2x^{2}-4x+2$

Subtract x from both sides of the equation:

$2x^{2}-5x+2=0$

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

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

Plug the first value of x into the second equation: y=-(2)+1=-1

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

x=2,y=-1 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:

$(2)=2(-1)^{2}=2$

(-1)=-(2)+1=-2+1=-1

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

$(\frac{1}{2})=2(\frac{1}{2})^{2}=2(\frac{1}{4})=\frac{1}{2}$

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

Solution:

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

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College Algebra : Systems of Equations

Study concepts, example questions & explanations for College Algebra

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Example Questions

Example Question #2 : Augmented Matrices

Example Question #7 : Augmented Matrices

Example Question #7 : Augmented Matrices

Example Question #1 : Non Linear Systems

Example Question #2 : Non Linear Systems

Example Question #3 : Non Linear Systems

Example Question #2 : Non Linear Systems

Example Question #3 : Non Linear Systems

Example Question #2 : Non Linear Systems

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