Download Solving Systems of Equations and Inequalities

Solving Systems of Equations
There are four ways of solving systems of equations: graphing, substitution, linear combinations (also called
elimination) and matrices (which are dealt with in another packet).
Graphing
Visually, this is the easiest to understand for many people. The idea is that there are two lines in a graph, and
the point where they cross is the solution. If the equations are not in y = mx + b form, you can use algebra to
put them in this form. It is not okay to just find the x or the y. You need both, because a point has two
dimensions, with a vertical and a horizontal component (or part). Remember: m = slope and b = y-intercept.
Example:
3
and 𝑏 = −1
1
2
𝑦 = −2𝑥 + 4 so 𝑚 = − and 𝑏 = 4
1
Now we graph these two lines and note where they cross.
𝑦 = 3𝑥 − 1 so 𝑚 =
As you can see, the lines cross at the point (1, 2), so this is the solution.
Example:
2
𝑦 = 2𝑥 + 3 so 𝑚 = and 𝑏 = 3
1
6
6
𝑦 = − 𝑥 + 3 so 𝑚 = − and 𝑏 = 3
5
5
The two lines cross at the point (0, 3), so that is the solution.
Example:
2x + y = -1
3y = -6x + 9
First change both to y = mx + b form.
y = -2x – 1
y = -2x + 3
Note that on these two equations, the slope is the same, but the y-intercept is different. These lines are parallel
and have no solution.
2
𝑥+2
3
2
𝑦= 𝑥+2
3
Example: − 2𝑥 + 3𝑦 = 6 ⟹ 3𝑦 = 2𝑥 + 6 ⟹ 𝑦 =
6𝑦 = 4𝑥 + 12 ⟹
2
x  2 . This means they are
3
the same line, and there are infinitely many solutions. I made the line purple, since red combines with blue to
make purple.
By rewriting both of these in y = mx + b form we see that both equations are y 

Substitution
This is exactly what it sounds like. You substitute one value or expression into the equation in place of one of
the letters (either x or y in these examples) to eliminate one of your variables.
Many students find this quite difficult, because they don’t understand that the expression y = 3x + 4 means that
you can either say y or you can say 3x + 4, and the two statements are exactly the same. One way of looking at
this is with naming people. You can meet a person named Andreas Michael Johnson. You may call him
Andreas, Mr. Johnson, Andy, Mike, or any number of different names. However, you are still talking to the
same person. Andreas has not changed who he is. Only his name is altered.
It works the same way in math.
Example: 𝑦 = 2𝑥
𝑦 = 3𝑥 − 2
Notice that y equals both 2x and 3x – 2. Using the above example, Andreas Michael Johnson = Andreas and
Andreas Michael Johnson = Mr. Johnson. Since both Andreas and Mr. Johnson equal Andreas Michael
Johnson, Andreas = Mr. Johnson. They are the same person. Similarly,
2x = 3x – 2
By substituting 2x in for y in the second equation, we have eliminated y as one of our variables, and we can now
solve the problem for x. Another way of looking at this is purple = red and purple = blue, so red = blue.
2𝑥
−3𝑥
−𝑥
𝑥
= 3𝑥 − 2
= −3𝑥
= −2
=2
Notice that I subtract 3𝑥 from both sides using an "adding line. "
Now that we know x = 2, we can substitute 2 back in for x in one of the original equations. I will choose y = 2x,
but it does not matter.
So 𝑦 = 2𝑥
𝑦 = 2(2)
𝑦=4
The solution is the point (2, 4).
Note that you need both the x and the y
parts of this answer. The x part tells you
how far to go left or right, and the y part
tells you how far to go up and down on a
graph.
This graph of the two lines shows that
the algebraic solution works.
Example:
3x + y = 3
9x + 2y = 7
Solve for one of the variables in one of the equations. That means put y (or x) on one side of the equals sign and
everything else on the other side. I will solve the first equation for y because y is already by itself.
3x + y = 3
-3x
-3x
y = -3x + 3
Substitute this new value for y into the second equation. Every time you see a y in the second equation, replace
it with –3x + 3. Common error: You cannot combine -3x and 3.
9x + 2y = 7
9x + 2(– 3x + 3) = 7
Note the parentheses. They must be there.
Solve as usual.
9𝑥 + (−6𝑥) + 6 = 7
3𝑥 + 6 = 7
−6 = −6
3𝑥 = 1
1
𝑥=
3
1
back in for x in one of the original equations and figure out y.
3
9𝑥 + 2𝑦 = 7
1
9 ( ) + 2𝑦 = 7
3

3 + 2𝑦 = 7
2𝑦 = 4
𝑦=2
1
The solution is ( , 2).
3
Substitute
Linear Combinations (also called Elimination)
Linear combinations just means that we add or subtract our two equations to get rid of (eliminate) a variable.
Example:
5x + y = 16
-5x + 3y = 8
0x + 4y = 24
4y = 24
y=6
Add these equations together. Note: y = 1y.
Solve for y.
Substitute 6 in for y in either one of the original equations and find x. The only difference will be in the
difficulty of solving the equation.
5x + 6 = 16
5x = 10
x=2
or
-5x + 3(6) = 8
-5x + 18 = 8
-5x = -10
x=2
The solution is (2, 6).
Example:
7x + 16y = 9
10x + 16y = 6
-3x + 0y = 3
-3x = 3
x = -1
Subtract these equations.
Substitute –1 in for x in one of the original equations and solve for y.
7(-1) + 16y = 9
-7 + 16y = 9
16y = 16
y=1
The solution is (-1, 1).
or
10(-1) + 16y = 6
-10 + 16y = 6
16y = 16
y=1
Sometimes the problem is not lined up nice and neat. You may need to multiply one or both equations by the
LCM (least common multiple). This is done the same way as finding a common denominator when adding
fractions.
Example:
4x – 3y = 17
5x + 4y = 60
This one has no obvious variable to solve for, so multiply both equations by the LCM of the coefficients of one
of the variables (I choose to eliminate x here. It does not matter which variable you choose). The LCM of 4
and 5 (the coefficients of x) is 20, so I multiply the first equation by 5 and the second equation by -4.
5(4x – 3y = 17)
-4(5x + 4y = 60)
⟹
⟹
20x – 15y = 85
-20x – 16y = -240
0x – 31y = -155
y=5
Solve for x.
5x + 4(5) = 60
5x + 20 = 60
5x = 40
x=8
The solution is (8, 5).
or
4x – 3(5) = 17
4x – 15 = 17
4x = 32
x=8
Add these.
Special Cases
Occasionally you get an odd problem with no solution, or the two lines coincide (are the same line). How can
you tell? They are the same line if you solve the problem using one of the above methods, and you get an
answer like 5 = 5 instead of x = 5. Another way to tell is by using the linear combinations method. When you
multiply through by the LCM, the two lines have the same equation. A similar problem has been done in the
graphing section.
Example (using substitution): 2x + y = 7
6x + 3y = 21
Change the first equation to y = -2x + 7 and substitute into the second equation.
6x + 3(-2x + 7) = 21
6x – 6x + 21 = 21
0 + 21 = 21
21 = 21
Alternatively (using linear combinations):
2x + y = 7
⟹
3(2x + y = 7) ⟹
6x + 3y = 21 ⟹
6x + 3y = 21 ⟹
6x + 3y = 21
6x + 3y = 21
Note that they are the same equation.
Two lines are parallel (no solution) if they have the same slope but different y-intercepts. If you use
substitution or linear combinations you will get a nonsense answer like 5 = 8. A similar problem has been done
in the graphing section.
Example:
3x + y = 4
6x + 2y = -2
Using substitution:
y = -3x + 4
(put the first equation in slope-intercept form)
So 6x + 2(-3x + 4) = -2 (substitute into the second equation)
6x – 6x + 8 = -2
0 + 8 = -2
8 = -2
Using linear combinations:
3x + y = 4
6x + 2y = -2
→
→
So again, you have a nonsense answer.
2(3x + y = 4) →
6x + 2y = -2 →
6x + 2y = 8
6x + 2y = -2
0 = 10
Systems of Linear Inequalities
I’m going to use the same examples I did for Systems of Linear Equations so you can compare them and see
how they work. The main thing to notice is that the lines graph exactly the same. The only differences are in
deciding whether the line is dashed or solid, and then deciding which side of the line to shade.
Example:
3
and 𝑏 = −1
1
2
𝑦 ≤ −2𝑥 + 4
notice that 𝑚 = − and 𝑏 = 4
1
Now we graph these two lines as if they were using regular = signs instead of the inequality symbols.
𝑦 < 3𝑥 − 1
notice that 𝑚 =
The next step is to decide whether the lines are dashed or solid. The first equation (red) is a plain “less than” so
the line will be dashed. The second equation (blue) is a “greater than or equal to” so the line will be solid.
The third and final step is to pick a test point and decide if it is true or false for each inequality. I like to use the
point (0, 0) whenever possible because it is very easy to substitute into each of the inequalities. If that point is
on one or both of the lines, you need to choose a different test point. Obvious choices include (1, 0) and (0, 1).
With only two lines, one of the three points listed here will be available.
The equations are 𝑦 < 3𝑥 − 1 and 𝑦 ≤ −2𝑥 + 4.
Since (0, 0) is not on either line, I can use it as my test point for both inequalities.
𝑦 < 3𝑥 − 1
0 < 3(0) − 1
0 < −1
This is FALSE, so shade the side of this line that does NOT include the point (0, 0) (red and purple area).
𝑦 ≤ −2𝑥 + 4
0 ≤ −2(0) + 4
0≤4
This is TRUE, so shade the side of this line that DOES include the point (0, 0) (blue and purple area).
The triangular region below and right of the dashed line and below and left of the solid line, where the red and
blue areas overlap is the answer (purple area).
One thing many people find difficult to understand it that the answer to one of these problems isn’t a single
point. It is a large group of points that fit anywhere inside the purple region. For example, (2, -9) and (-2, -50)
are both solutions to this problem. If you are not sure, try these two points in both original equations and you
will see that you get a true answer.
Example:
𝑦 ≥ 2𝑥 + 3
6
𝑦 <− 𝑥+3
5
2
and 𝑏 = 3
1
6
here 𝑚 = − 𝑥 and 𝑏 = 3
5
here 𝑚 =
The lines graph just like they did in the original problem on page 2. Since the first line is “greater than or equal
to” it has a solid line. Since the second equation is “less than or equal to” it has a dashed line.
Again, (0, 0) will work as a test point for both lines, since it is not on either line.
6
𝑦 ≥ 2𝑥 + 3
𝑦 <− 𝑥+3
5
6
0 ≥ 2(0) + 3
0 < − (0) + 3
5
0≥3
False (red)
0<3
True (blue)
The area where red and blue overlap (purple) is the answer. Anything to the left of the solid red line and to the
left and below the blue dashed line will work. Sample points that are solutions to this system of linear
inequalities include (-10, 8) and (-150, -6). There are infinitely many solutions, which is why the region is
shaded, rather than naming all the specific points.
Example: 2𝑥 + 𝑦 ≥ −1
3𝑦 ≤ −6𝑥 + 9
First, change both equations to slope-intercept form (𝑦 = 𝑚𝑥 + 𝑏).
𝑦 = −2𝑥 − 1
𝑦 = −2𝑥 + 3
Note that on these two equations, the slope is the same, but the y-intercept is different. These lines are parallel.
Since both lines have the “or equal to” symbol, both are solid lines.
𝑦 ≥ −2𝑥 − 1
𝑦 ≤ −2𝑥 + 3
Now we pick a test point. (0, 0) will work again, since neither line goes through the origin, but just so you can
see how any point will work, I will choose something different.
For the first equation, I will choose the point (2, 4) and mark it on the graph in red.
For the second equation, I will choose the point (6, 3) and mark it on the graph in blue.
Now we test the two points.
𝑦 ≥ −2𝑥 − 1 ⟹ 4 ≥ −2(2) − 1 ⟹ 4 ≥ −4 − 1 ⟹ 4 ≥ −5 𝑻𝒓𝒖𝒆
𝑦 ≤ −2𝑥 + 3 ⟹ 3 ≤ −2(6) + 3 ⟹ 3 ≤ −12 + 3 ⟹ 3 ≤ −8 𝑭𝒂𝒍𝒔𝒆
The red point works and the blue point does not. So I will shade the side of the red line that includes the point
(2,4) and I will shade the side of the blue line that does NOT include (1, 3).
The answer is the purple area between the red and blue lines. It is a REGION, not a point. ANY point inside
the purple area will work to answer this question.
Practice Problems
1. Find the intersection of this system of equations by graphing. Do not show any tables.
3𝑥 + 2𝑦 = 12
𝑥 − 2𝑦 = −4
The intersection is (
,
)
Solve the following systems of equations using substitution or linear combinations (algebra). Show ALL work
for full credit.
2. 𝑦 = 2𝑥
𝑦 = −2𝑥 − 4
(
,
)
4. 𝑥 = 9
𝑦 = 4𝑥 − 12
(
,
)
6. 2𝑥 − 5𝑦 = 17
6𝑥 − 5𝑦 = 1
(
,
)
3. 𝑥 + 𝑦 = −4
− 2𝑥 + 𝑦 = 5
(
,
)
5. 3𝑥 − 2𝑦 = 8
𝑥 + 2𝑦 = 8
(
,
)
7. 2𝑥 + 4𝑦 = 21
3𝑥 + 7 = 11
(
,
)
Key
1
3
1. 𝑦 = 𝑥 + 2 and 𝑦 = − 𝑥 + 6
2
2
with intersection at (2, 3)
y
5
#1
4
3
2. (-1, -2)
3. (-3, -1)
4. (9, 24)
5. (4, 2)
6. (-4, -5)
23 41
7. ( , )
10 10
2
1
-5
-4
-3
-2
-1
-2
-3
-4
-5
x
1
2
3
4
5