Solving Linear Systems Algebraically Using Addition/Subtraction Topic Index | Algebra Index | Regents Exam Prep Center

Solve this system of equations using the addition or subtraction method.  Check.

x - 2y = 14
x + 3y = 9

 Simultaneous equations got you baffled?   Relax!  You can do it!  Think of the adding or subtracting 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 "addition" or "subtraction" 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!)

 Grab you Graphing Calculator: Even though you are doing an "algebraic solution" you can still use your graphing calculator as a CHECK to be sure you have the correct answer.  Click the calculator at the right for directions on using the TI-83+/84+ graphing calculator to solve systems of equations.                                                                             Click calculator.

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