Linear Systems with Two Variables - College Algebra

Card 0 of 64

Question

If

and

$x=\frac{2}{3}y-2$

Solve for and .

Answer

rearranges to

and

$x=\frac{2}{3}y-2$ , so

$32-y=\frac{2}{3}y-2$

$32-1\frac{2}{3}y=-2$

$-1\frac{2}{3}y=-34$

$-\frac{5}{3}y=-34$

$y=\frac{-34}{1}\cdot\frac{-3}{5}=\frac{102}{5}=20\frac{2}{5}$

$x=32-20\frac{2}{5}=11\frac{3}{5}$

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Question

Solve for in the system of equations:

Answer

In the second equation, you can substitute for from the first.

Now, substitute 2 for in the first equation:

The solution is

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Question

Solve the system of equations.

Answer

Isolate in the first equation.

Plug into the second equation to solve for .

Plug into the first equation to solve for .

Now we have both the and values and can express them as a point: .

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Question

Solve for and .

Answer

1st equation:

2nd equation:

Subtract the 2nd equation from the 1st equation to eliminate the "2y" from both equations and get an answer for x:

Plug the value of into either equation and solve for :

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Question

What is a solution to this system of equations?

$\left{\begin{matrix} 2y=3x+11\ y=-5x-14\end{matrix}\right.$

Answer

Substitute equation 2. into equation 1.,

$2\left(-5x-14\right)=3x+11$

so,

Substitute into equation 2:

$y=-5\cdot (-3)-14=1$

so, the solution is .

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Question

A man in a canoe travels upstream 400 meters in 2 hours. In the same canoe, that man travels downstream 600 meters in 2 hours.

What is the speed of the current, , and what is the speed of the boat in still water, ?

Answer

This problem is a system of equations, and uses the equation $\text{distance}=\text{rate}\times\text{time}$ .

Start by assigning variables. Let stand for the rate of the boat, let stand for the rate of the current.

When the boat is going upstream, the total rate is equal to . You must subtract because the rates are working against each other—the boat is going slower than it would because it has to work against the current.

Using our upstream distance (400m) and time (2hr) from the question, we can set up our rate equation:

$400 = (b-c) \times 2$

When the boat is going downstream, the total rate is equal to because the boat and current are working with each other, causing the boat to travel faster.

We can refer to the downstream distance (600m) and time (2hr) to set up the second equation:

$600 = (b+c) \cdot 2$

From here, use elimination to solve for and .

1. Set up the system of equations, and solve for .

2. Subsitute into one of the equations to solve for .

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Question

Nick’s sister Sarah is three times as old as him, and in two years will be twice as old as he is then. How old are they now?

Answer

Step 1: Set up the equations

Let = Nick's age now
Let = Sarah's age now

The first part of the question says "Nick's sister is three times as old as him". This means:

The second part of the equation says "in two years, she will be twice as old as he is then). This means:

Add 2 to each of the variables because each of them will be two years older than they are now.

Step 2: Solve the system of equations using substitution

Substitute for in the second equation. Solve for

${\color{Red} n=2}$

Plug into the first equation to find

${\color{Red} s=6}$

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Question

Solve the system of equations:

Answer

Solve using elimination:

multiply the 2nd equation by two to make elimination possible

subtract 2nd equation from the first to solve for

${\color{Red} y=0}$

Substitute into either equation to solve for

${\color{Red} x=3}$

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Question

Solve for and :

Answer

There are two ways to solve this:

-The 1st equation can be mutliplied by while the 2nd equation can be multiplied by and added to the 1st equation to make it a single variable equation where

.

This can be plugged into either equation to get

or

-The 2nd equation can be simplified to,

.

This value for can then be substituted into the first equation to make the equation single variable in .

Solving, gives , which can be plugged into either original equation to get

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Question

Solve the system of equations.

Answer

Use elimination, multiply the top equation by -4 so that you can eliminate the X's.

__________________

Combine these two equations, and then you have;

Plug in y into one of the original equations and solve for x.

Your solution is .

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Question

Solve the system of equation using elimination:

$\x+3y=2\4x+y=19$

Answer

To solve by elimination, we want to cancel out either the or variable:

$\-4(x+3y)=-4(2)\4x+y=19$

$\-4x-12y=-8\4x+y=19$

$\-12y=-8\y=19$

$\-11y=11\y=-1$

Now that we know the value of , we can plug it in to one of the equation sets to solve for :

$\x+3(-1)=2\x-3=2\x=5$

We can then conclude that

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Question

Solve this linear system:

Answer

To solve this linear system we must find the coordinate point where both lines intersect. To do so, we eliminate one variable (by multiplying or dividing by some amount) and solve for the remaining variable.

Multiply equation 1 by 5 and equation 2 by 6; add the two equations to cancel out the x term:

Solve for y:

Now substitute this value back into either original equation to get the y-coordinate:

$6x+18.9=10\rightarrow6x=-8.9\rightarrow x=-1.5$

Rounded answer: $\boldsymbol{(-1.5,6.3)}$

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Question

Solve the following system of equations:

Answer

Solve by adding the two equations:

Adding the two equations yields the following:

Dividing both sides of the equation by 3 yields the following:

Plug value of x back into first equation:

Subtracting 3 from both sides of the equation yields the following:

Make sure the values for x and y satisfy both equations:

Solution:

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Question

Solve for : $\begin{bmatrix} 2x+5y =1\ 3x+5y =4 \end{bmatrix}$

Answer

We can evaluate the value of by subtracting the first equation from the second equation since both equations share , and can be eliminated.

The equation becomes:

Substitute this value back into either the first or second equation, and solve for y.

The answer is:

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Question

Consider the system of linear equations

True or false: This system has one and only one solution.

Answer

The given system has more equations than variables, 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

$\begin{bmatrix} 1 & 3 & 20 \ 2 & 6 & 40 \ -3 & 9 & 60 \end{bmatrix}$

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 operations:

$-2R1+R2 \rightarrow R2$

$3R1+R3 \rightarrow R3$

$\begin{bmatrix} 1 & 3 & 20 \ -2(1)+ 2 & -2(3)+ 6 & -2(20)+40 \ 3(1)+ (-3) &3(3)+ 9 &3(20)+ 60 \end{bmatrix}$

$\begin{bmatrix} 1 & 3 & 20 \ 0 & 0 & 0 \ 0&18&120 \end{bmatrix}$

An all zero row has been created, so it must be moved to the bottom:

$R2 \leftrightarrow R3$

$\begin{bmatrix} 1 & 3 & 20\ 0&18&120 \ 0 & 0 & 0 \end{bmatrix}$

Now, get a 1 into Row 2, Column 2:

$\frac{1}{18}R2 \rightarrow R2$

$\begin{bmatrix} 1 & 3 & 20\ \frac{1}{18} (0)&\frac{1}{18} (18)&\frac{1}{18} (120) \ 0 & 0 & 0 \end{bmatrix}$

$\begin{bmatrix} 1 & 3 & 20\ 0&1&\frac{20}{3}\ 0 & 0 & 0 \end{bmatrix}$

Now, get 0's in the other positions in Column 2:

$-3R2+ R1 \rightarrow R1$

$\begin{bmatrix}-3(0)+ 1 &-3(1)+ 3 & -3 ( \frac{20}{3})+20\ 0&1&\frac{20}{3}\ 0 & 0 & 0 \end{bmatrix}$

$\begin{bmatrix} 1 & 0 & 0\ 0&1&\frac{20}{3}\ 0 & 0 & 0 \end{bmatrix}$

This matrix is in reduced row-echelon form, and can be interpreted to mean that

$x = 0, y = \frac{20}{3}$ .

Thus, the system of equations given has one and only one solution.

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Question

Solve this system of equations:

Answer

To solve this system of equations we must first eliminate one variable and solve for the remaining variable. We then substitute the variable back into our original equation and solve for the second variable still unknown.

We will use the elimination method:

Multiply the top equation by 1 and add it to the second equation:

--------------------------

$x=-\frac{6}{5}$

Now we substitute the value of x into our original equation:

$2(\frac{-6}{5})+5y=15\rightarrow 5y=\frac{12}{5}+15\rightarrow y=\frac{87}{25}$

Thus, our lines are equal (intersect) at $\boldsymbol{(\frac{-6}{5},\frac{87}{25})}$ .

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Question

If

and

$x=\frac{2}{3}y-2$

Solve for and .

Answer

rearranges to

and

$x=\frac{2}{3}y-2$ , so

$32-y=\frac{2}{3}y-2$

$32-1\frac{2}{3}y=-2$

$-1\frac{2}{3}y=-34$

$-\frac{5}{3}y=-34$

$y=\frac{-34}{1}\cdot\frac{-3}{5}=\frac{102}{5}=20\frac{2}{5}$

$x=32-20\frac{2}{5}=11\frac{3}{5}$

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Question

Solve for in the system of equations:

Answer

In the second equation, you can substitute for from the first.

Now, substitute 2 for in the first equation:

The solution is

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Question

Solve the system of equations.

Answer

Isolate in the first equation.

Plug into the second equation to solve for .

Plug into the first equation to solve for .

Now we have both the and values and can express them as a point: .

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Question

Solve for and .

Answer

1st equation:

2nd equation:

Subtract the 2nd equation from the 1st equation to eliminate the "2y" from both equations and get an answer for x:

Plug the value of into either equation and solve for :

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