Intermediate Algebra
Answer/Discussion to Practice Problems
Tutorial 19: Solving Systems of Linear Equations
in Two Variables

WTAMU > Virtual Math Lab > Intermediate Algebra > Tutorial 19: Solving Systems of Linear Equations

Answer/Discussion to 1a

problem 1a

Step 1: Graph the first equation.

x-intercept

*Plug in 0 for y for x-int

*Inverse of mult. by 5 is div. by 5

*x-intercept

The x-intercept is (2, 0).

y-intercept

*Plug in 0 for x for y-int

*Inverse of mult. by -2 is div. by -2

*y-intercept

The y-intercept is (0, -5).

Find another solution by letting x = 1.

*Plug in 1 for x
*Inverse of add 5 is sub. 5

*Inverse of mult. by -2 is div. by -2

Another solution is (1, -5/2).

Solutions:

x	y	(x, y)
2	0	(2, 0)
0	-5	(0, -5)
1	-5/2	(1, -5/2)

Plotting the ordered pair solutions and drawing the line:

ad1a4

Step 2: Graph the second equation on the same coordinate system as the first.

x-intercept

*Plug in 0 for y for x-int
*x-intercept

The x-intercept is (-1, 0).

y-intercept

*Plug in 0 for x for y-int

*Inverse of mult. by -1 is div. by -1

*y-intercept

The y-intercept is (0, 1).

Find another solution by letting x = 1.

*Plug in 1 for x
*Inverse of add 1 is sub. 1

*Inverse of mult. by -1 is div by -1

Another solution is (1, 2).

Solutions:

x	y	(x, y)
-1	0	(-1, 0)
0	1	(0, 1)
1	2	(1, 2)

Plotting the ordered pair solutions and drawing the line:

ad1a8

Step 3: Find the solution.

We need to ask ourselves, is there any place that the two lines intersect, and if so, where?

The answer is yes, they intersect at (4, 5).

Step 4: Check the proposed ordered pair solution in BOTH equations.

You will find that if you plug the ordered pair (4, 5) into BOTH equations of the original system, that this is a solution to BOTH of them.

The solution to this system is (4, 5).

(return to problem 1a)

Answer/Discussion to 2a

problem 2a

Step 1: Simplify if needed.

Both of these equations are already simplified. No work needs to be done here.

Step 2: Solve one equation for either variable.

Neither equation is already solved for either variable. It looks like equation one is the easiest one to work with. Either x or y would be fine to solve for in the first equation since they both have a 1 as a coefficient.

I will choose to solve for x in the first equation:
(Note it would be perfectly fine to solve for y)

*Inverse of add y is sub. y
*Solved for x

Step 3: Substitute what you get for step 2 into the other equation
AND
Step 4: Solve for the remaining variable.

Substitute the expression 3 - y for x into the second equation and solve for y:
(when you plug in an expression like this, it is just like you plug in a number for your variable)

*Sub. 3 - y in for x
*Dist. 2 through ( )
*Combine like terms

*Inverse of add 6 is sub. 6

Step 5: Solve for second variable.

Plug in -10 for y into the equation in step 2 to find x's value.

*Plug in -10 for y

Step 6: Check the proposed ordered pair solution in BOTH original equations.

You will find that if you plug the ordered pair (13, -10) into BOTH equations of the original system, that this is a solution to BOTH of them.

(13, -10) is a solution to our system.

(return to problem 2a)

Answer/Discussion to 3a

problem 3a

Step 1: Simplify and put both equations in the form Ax + By = C if needed.

Since we have some fractions in this problem, let's go ahead and get rid of them to simplify things.

Multiplying first equation by it's LCD we get:

*Mult. by LCD of 2

Step 2: Multiply one or both equations by a number that will create opposite coefficients for either x or y if needed.

We can either work with the x's or the y's; it doesn't matter.

I'm going to choose to work with the x's. The smallest number that both 6 and 9 go into is 18. This works in the same fashion as LCD, you want the smallest number they both go into. Make one of them positive and the other negative (it doesn't matter which one is which), so that you have opposites.

Multiplying the first equation by 3 and the second equation by -2 we get: