Linear Systems with Three Variables - College Algebra

Card 0 of 7

Question

Solve this system of equations.

Answer

Equation 1:

Equation 2:

Equation 3:

Adding the terms of the first and second equations together will yield .

Then, add that to the third equation so that the y and z terms are eliminated. You will get .

This tells us that x = 1. Plug this x = 1 back into the systems of equations.

Now, we can do the rest of the problem by using the substitution method. We'll take the third equation and use it to solve for y.

Plug this y-equation into the first equation (or second equation; it doesn't matter) to solve for z.

$z=-\frac{5}{3}$

We can use this z value to find y

$y=-3\left ( -\frac{5}{3}\right )-3$

So the solution set is x = 1, y = 2, and z = –5/3.

Compare your answer with the correct one above

Question

What is a solution to this system of equations:

$\left{\begin{matrix} x+2y-z=6\ 2x+y+z=9\ 3x+4y+z=18\end{matrix}\right.$

Answer

Step 1: Multiply first equation by −2 and add the result to the second equation. The result is:

$\left{\begin{matrix} x+2y-z=6\ -3y+3z=-3\ 3x+4y+z=18\end{matrix}\right.$

Step 2: Multiply first equation by −3 and add the result to the third equation. The result is:

$\left{\begin{matrix} x+2y-z=6\ -3y+3z=-3\ -2y+4z=0\end{matrix}\right.$

Step 3: Multiply second equation by −23 and add the result to the third equation. The result is:

$\left{\begin{matrix} x+2y-z=6\ -3y+3z=-3\ 2z=2\end{matrix}\right.$

Step 4: solve for z.

Step 5: solve for y.

$-3y+3\times1=-3$

Step 6: solve for x by substituting y=2 and z=1 into the first equation.

$x+2\times2-1=6$

Compare your answer with the correct one above

Question

Solve the equations for , , and .

Answer

To solve the system of linear equations, first isolate the equations with and to help solve for . This will include substitution.

Now that is known it can be substituted back into the third equation to solve for .

Lastly, substitute into the first equation and solve for .

Therefore, the solution to this system is

Compare your answer with the correct one above

Question

Solve the following system of equations for a, b, and c.

Answer

To solve the equations we want to try to cancel out some of the letters to reduce the equations. We can do this by adding different equations to each other. If we add:

we get a new equation,

.

We can also add

to get .

So we can rearrange the new equations to get

and substitute these into the original second line equation:

$a+(4a-7)-(13-2a)=1\rightarrow 7a -20 = 1$

$7a = 21 \rightarrow a=3$

Then we can solve for b and c:

$b= 4\cdot 3-7 = 5$

$c=13-2\cdot 3 = 7$

Compare your answer with the correct one above

Question

Solve the following system of equations:

Answer

Add the first and third equations together, to eliminate z:

+

This yields the following:

Divide out a 2, to simplify the equation:

Plug this value back into the first equation:

Solve for z:

Add the first and second equations together to eliminate y:

+

This yields the following:

Divide out a 3 to simplify the equation:

Plug in the value for z:

Solve for x:

Plug the values for x and z into the first equation:

Simplify:

Solve for y:

Make sure the values for x,y, and z satisfy all of the equations:

Solution:

Compare your answer with the correct one above

Question

Determine the value of y: $\begin{bmatrix} x+y+z=1\ -2x-y-z =3\x+y+3z = -1 \end{bmatrix}$

Answer

Add the first two equations to eliminate the y and z-variable.

$\begin{bmatrix} x+y+z=1\ -2x-y-z =3 \end{bmatrix} \rightarrow -x = 4 \rightarrow x=-4$

Using the value of x, with the first and third equations, we will need to eliminate the z-variable to solve for y.

$\begin{bmatrix} (-4)+y+z=1\(-4)+y+3z = -1 \end{bmatrix}\rightarrow \begin{bmatrix} y+z=5\y+3z = 3 \end{bmatrix}$

Multiply the top equation by three.

$\begin{bmatrix} 3y+3z=15\y+3z = 3 \end{bmatrix}$

Subtract both equations.

Divide both sides by two.

The answer is:

Compare your answer with the correct one above

Question

Consider the system of linear equations

Describe the system.

Answer

The given system has as many variables as equations, which makes it possible to have exactly one solution.

One way to identify the solution set is to use Gauss-Jordan elimination on the augmented coefficient matrix

$\begin{bmatrix} 1 & 2 & 0 & 8 \ 0 & 1 & 2 & 10 \ 1 & 0 & 2 & 12 \end{bmatrix}$

Perform operations on the rows, with the object of rendering this matrix in reduced row-echelon form.

First, a 1 is wanted in Row 1, Column 1. This is already the case, so 0's are wanted elsewhere in Column 1. Do this using the row operation:

$-R1+R3 \rightarrow R3$

$\begin{bmatrix} 1 & 2 & 0 & 8 \ 0 & 1 & 2 & 10 \ -1+ 1 &-2+ 0 & -0+2 &-8+ 12 \end{bmatrix}$

$\begin{bmatrix} 1 & 2 & 0 & 8 \ 0 & 1 & 2 & 10 \ 0 & -2 & 2 & 4\end{bmatrix}$

Next, a 1 is wanted in Row 2, Column 2. This is already the case, so 0's are wanted elsewhere in Column 2. Do this using the row operations:

$\begin{bmatrix} 1 & -2(1)+2 & -2(2)+0 & -2(10)+8 \ 0 & 1 & 2 & 10 \ 2(0)+0 &2(1)+ (-2 )&2(2)+ 2 &2(10)+ 4\end{bmatrix}$

$\begin{bmatrix} 1 & 0 & -4 & -12\ 0 & 1 & 2 & 10 \ 0 & 0 & 6 &2 4\end{bmatrix}$

Next, a 1 is wanted in Row 3, Column 3.Perform the operation

$\frac{1}{6} R3 \rightarrow R3$

$\begin{bmatrix} 1 & 0 & -4 & -12\ 0 & 1 & 2 & 10 \ \frac{1}{6}(0 )& \frac{1}{6}(0) & \frac{1}{6}(6) &\frac{1}{6}(2 4)\end{bmatrix}$

$\begin{bmatrix} 1 & 0 & -4 & -12\ 0 & 1 & 2 & 10 \ 0 & 0 & 1 & 4\end{bmatrix}$

Now, 0's are wanted in the rest of Column 3:

$4R3 + R1 \rightarrow R1$

$-2R3 + R2 \rightarrow R2$

$\begin{bmatrix} 4(0)+1 &4(0)+ 0 & 4(1)+ (-4) & 4(4)+ (-12)\ -2(0)+0 &-2(0)+ 1 &-2(1)+ 2 & -2(4)+10 \ 0 & 0 & 1 & 4\end{bmatrix}$

$\begin{bmatrix} 1 & 0 & 0& 4\ 0 & 1 & 0 & 2 \ 0 & 0 & 1 & 4\end{bmatrix}$

The matrix is in the desired form and can be interpreted to mean that the system has one and only one solution - . This makes the system consistent and independent.

Compare your answer with the correct one above

Tap the card to reveal the answer