The following examples are from RegentsPrep.org
Practice Problems:
Systems of Equations
Systems of Equations Word Problems
Systems of Inequalities

The substitution method is used to eliminate one of the variables by replacement when solving a system of equations.
Think of it as “grabbing” what one variable equals from one equation and “plugging” it into the other equation.


Systems of Equations may also be referred to as “simultaneous equations”.
Let’s look at an example using the substitution method:
Solve this system of equations
(and check): 

3y – 2x = 11
y + 2x = 9 
1. Solve one of the equations for either “x =” or “y =”.
This example solves the second equation for “y =”. 

3y – 2x = 11
y = 9 – 2x 
2. Replace the “y” value in the first equation by what “y” now equals. Grab the “y” value and plug it into the other equation. 

3(9 – 2x) – 2x = 11 
3. Solve this new equation for “x“. 

(27 – 6x) – 2x = 11
27 – 6x – 2x = 11
27 – 8x = 11
8x = 16
x = 2 
4. Place this new “x” value into either of the ORIGINAL equations in order to solve for “y“. Pick the easier one to work with! 

y + 2x = 9 or
y = 9 – 2x
y = 9 – 2(2)
y = 9 – 4
y = 5 
5. Check: substitute x = 2 and y = 5 into BOTH ORIGINAL equations. If these answers are correct, BOTH equations will be TRUE! 

3y – 2x = 11
3(5) – 2(2) = 11
15 – 4 = 11
11 = 11 (check!) 
y + 2x = 9
5 + 2(2) = 9
5 + 4 = 9
9 = 9 (check!)

The Elimination Method

Simultaneous equations got you baffled? Relax! You can do it!
Think of the elimination method as temporarily “eliminating” one of the variables to make your life easier.


Systems of Equations may also be referred to as “simultaneous equations”.
“Simultaneous” means being solved “at the same time”.
Let’s look at three examples using the “elimination” method for systems of equations:
1. Solve this system of equations
and check: 

x – 2y = 14
x + 3y = 9 
a. First, be sure that the variables are “lined up” under one another. In this problem, they are already “lined up”. 

x – 2y = 14
x + 3y = 9 
b. Decide which variable (“x” or “y“) will be easier to eliminate. In order to eliminate a variable, the numbers in front of them (the coefficients) must be the same or negatives of one another. Looks like “x” is the easier variable to eliminate in this problem since the x‘s already have the same coefficients. 

x – 2y = 14
x + 3y = 9 
c. Now, in this problem we need to subtract to eliminate the “x” variable. Subtract ALL of the sets of lined up terms.
(Remember: when you subtract signed numbers, you change the signs and follow the rules for adding signed numbers.) 

x – 2y = 14
–x – 3y = – 9
– 5y = 5 
d. Solve this simple equation. 

5y = 5
y = 1 
e. Plug “y = 1″ into either of the ORIGINAL equations to get the value for “x“. 

x – 2y = 14
x – 2(1) = 14
x + 2 = 14
x = 12 
f. Check: substitute x = 12 and y = 1 into BOTH ORIGINAL equations. If these answers are correct, BOTH equations will be TRUE! 

x – 2y = 14
12 – 2(1) = 14
12 + 2 = 14
14 = 14 (check!)
x + 3y = 9
12 + 3(1) = 9
12 – 3 = 9
9 = 9 (check!) 


There is no stopping us now!
Let’s try a harder problem…. 
2. Solve this system of equations
and check: 

4x + 3y = 1
5x + 4y = 1 
a. You can probably see the dilemma with this problem right away. Neither of the variables have the same (or negative) coefficients to eliminate. Yeek! 

4x + 3y = 1
5x + 4y = 1 
b. In this type of situation, we must MAKE the coefficients the same (or negatives) by multiplication. You can MAKE either the “x” or the “y” coefficients the same. Pick the easier numbers. In this problem, the “y” variables will be changed to the same coefficient by multiplying the top equation by 4 and the bottom equation by 3.
Remember:
* you can multiply the two differing coefficients to obtain the new coefficient if you cannot think of another smaller value that will work.
* multiply EVERY element in each equation by your adjustment numbers. 

4(4x + 3y = 1)
3(5x + 4y = 1)
16x + 12y = 4
15x + 12y = 3

c. Now, in this problem we need to subtract to eliminate the “y” variable.
(Remember: when you subtract signed numbers, you change the signs and follow the rules for adding signed numbers.) 

16x + 12y = 4
–15x – 12y = – 3
x = – 7 
d. Plug “x = 7″ into either of the ORIGINAL equations to get the value for “y“. 

5x + 4y = 1
5(7) + 4y = 1
35 + 4y = 1
4y = 36
y = 9 
e. Check: substitute x = 7 and y = 9 into BOTH ORIGINAL equations. If these answers are correct, BOTH equations will be TRUE! 

4x + 3y = 1
4(7) +3(9) = 1
28 + 27 = 1
1 = 1 (check!)
5x + 4y = 1
5(7) + 4(9) = 1
35 + 36 = 1
1 = 1 (check!)


Let’s finish with an addition method problem:
3. Solve this system of equations and check: 

4x – y = 10
2x = 12 3y 
a. First, be sure that the variables are “lined up” under one another. The second equation was rearranged so that the variables would “line up” with those in the first equation. 

4x – y = 10
2x + 3y = 12 
b. Decide which variable (“x” or “y“) will be easier to eliminate. In this problem, we must MAKE EITHER the “x” or the “y” coefficients the same. The “y” variable is being used here. Multiplying by 3 will give the “y” variables negative coefficients. (Yes, 3 could also have been used.) 

3(4x – y = 10)
2x + 3y = 12
12x – 3y = 30
2x + 3y = 12 
c. Now, add to eliminate the “y” variable.


12x – 3y = 30
2x + 3y = 12
14x = 42 
d. Solve this simple equation. 

14x = 42
x = 3 
e. Plug “x = 3″ into either of the ORIGINAL equations to get the value for “y”. 

4x – y = 10
4(3) – y = 10
12 – y = 10
–y = 2
y = 2 
f. Check: substitute x = 3 and y = 2 into BOTH ORIGINAL equations. If these answers are correct, BOTH equations will be TRUE! 

4x – y = 10
4(3) – 2 = 10
12 – 2 = 10
10 = 10 (check!)
2x = 12 3y
2(3) = 12 – 3(2)
6 = 12 – 6
6 = 6 (check!) 
If you can graph an inequality, you can graph a system of inequalities!
Simply graph each inequality separately on the same set of axes. The area where the shadings overlap is the solution to the system of inequalities.


To Graph a System of Inequalities

1. Graph each inequality separately. The directions for graphing inequalities can be found in “Graphing Inequalities”.2. The solution to the system, will be the area where
the shadings from each inequality overlap one
another. 

Graph the following system of inequalities


Graph the system

Graph each inequality as if it was stated in “y=” form. If the inequality is < or > a dotted line must be used to represent the line. If the inequality is < or >, a solid line is used.
Choose a test point to determine which side of the line needs to be shaded. The test point for this problem was (0,0). Always pick a point that is easy to work with.
For the test point (0,0),
0 < 2(0)3 False
0 (2/3)0+2 False
Since both equations were false, shading occurred on the other side of the line, not covering the point.
The solution, S, is where the two shadings overlap one another.


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