Title  College Algebra
Answer/Discussion to Practice Problems
Tutorial 49: Solving Systems of
Linear Equations in Two Variables Answer/Discussion to 1a I’m going to chose to use the elimination by addition method, however it would be perfectly find for you to use the substitution method.  Either way the answer will be the same.

 This equation is full of those nasty fractions.  We can simplify both equations by multiplying each separate one by it’s LCD, just like you can do when you are working with one equation.  As long as you do the same thing to both sides of an equation, you keep the two sides equal to each other. Multiplying each equation by it's respective LCD we get: *Mult. by LCD of 6 *Mult. by LCD of 15

 Again, you want to make this as simple as possible.  Note how the coefficient on y in the first equation is 2 and in the second equation it is 3.  We need to have opposites, so if one of them is 6 and the other is -6, they would cancel each other out when we go to add them.  If we added them together the way they are now, we would end up with one equation and two variables, nothing would drop out.  And we would not be able to solve it. So I proposed that we multiply the first equation by 3 and the second equation by -2, this would create a 10 and a -10 in front of the y’s and we will have our opposites.   Multiplying the first equation by 3 and the second equation by -2 we get: *Mult. both sides of 1st eq. by 3 *Mult. both sides of 2nd eq. by -2   *y's have opposite coefficients

 Step 3: Add equations. *Note that y's dropped out You can choose any equation used in this problem to plug in the found x value. I choose to plug in 2 for x into the first simplified equation (found in step 1) to find y’s value. *Plug in 2 for x   *Inverse of add 6 is sub. 6 *Inverse of mult. by 2 is div. by 2

 You will find that if you plug the ordered pair (2, -1) into BOTH equations of the original system, that this is a solution to BOTH of them. (2, -1) is a solution to our system. Answer/Discussion to 1b I’m going to chose to use the substitution method, however it would be perfectly find for you to use the elimination by addition method.  Either way the answer will be the same.

 Step 1: Simplify if needed.

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

 Note how the first equation is already solved for y.  We can use that one for this step. It does not matter which equation or which variable you choose to solve for.  But it is to your advantage to keep it as simple as possible.  First equation solved for y: *Solved for y

 Substitute the expression 2x - 5 for y into the second equation and solve for x: (when you plug in an expression like this, it is just like you plug in a number for your variable) *Sub. 2x - 5 for y *Variable dropped out AND true

 Wait a minute, where did our variable go???? As mentioned above, if the variable drops out AND we have a TRUE statement, then when have an infinite number of solutions.  They end up being the same line.

 Step 5:  Solve for second variable.

 Since we did not get a value for x, there is nothing to plug in here.

 There is no value to plug in here. When they end up being the same equation, you have an infinite number of solutions.  You can write up your answer by writing out either equation to indicate that they are the same equation. Two ways to write the answer are {(x, y)|  y = 2x - 5} OR {(x, y) | 4x - 2y = 10}. Answer/Discussion to 1c I’m going to chose to use the elimination by addition method, however it would be perfectly find for you to use the substitution method.  Either way the answer will be the same.

 Both of these equations are already simplified and in the right form.  No work needs to be done here.

 Again, you want to make this as simple as possible.  Note how the coefficient on x in the first equation is 2 and in the second equation it is -4.  We need to have opposites, so if one of them is 4 and the other is -4, they would cancel each other out when we go to add them.  If we added them together the way they are now, we would end up with one equation and two variables, nothing would drop out.  And we would not be able to solve it. So I proposed that we multiply the first equation by 2, this would create a 4 and a -4 in front of the x’s and we will have our opposites.   Multiplying the first equation by 2 we get: *Mult. both sides of 1st eq. by 2     *x's and y's have opposite coefficients

 Step 3: Add equations. *Note that x's and y's both dropped out

 Wait a minute, where did our variable go???? As mentioned above if your variable drops out and you have a FALSE statement, then there is no solution.  If we were to graph these two, they would be parallel to each other.

 There is no value to plug in here.

 There is no value to plug in here. The answer is no solution. Answer/Discussion to 2a 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: 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: 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 2a)

Last revised on March 26, 2011 by Kim Seward.
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