The addition method of
solving systems of equations is also called the method of elimination.
This method is similar to the method you probably learned for solving
simple equations.
If you had the equation
"<span>x + 6 = 11</span>", you would
write "–6"
under either side of the equation, and then you'd "add down"
to get "<span>x =
5</span>" as the solution.
<span>x + 6
= 11
–6 –6
x = 5</span>
You'll do something similar
with the addition method.
<span>Solve the following
system using addition.<span>2x + y<span> = 9
3</span>x – y = 16</span>Note that, if I add down,
the y's
will cancel out. So I'll draw an "equals" bar under the system,
and add down:2<span>x + y = 9
3x – y<span> = 16
</span>5x = 25</span>Now I can divide through
to solve for <span>x = 5</span>, and then back-solve,
using either of the original equations, to find the value of y.
The first equation has smaller numbers, so I'll back-solve in that one:<span><span>2(5) + y = 9
10 + y = 9
y = –1</span>Then the solution
is <span>(x, y) = (5, –1)</span>.</span></span>
It doesn't matter which equation
you use for the backsolving; you'll get the same answer either way. If
I'd used the second equation, I'd have gotten:
<span>3(5) – y = 16
15 – y = 16
–y = 1
y = –1</span>
...which is the same result
as before.
<span>Solve the following
system using addition.<span>x – 2y<span> = –9
</span>x + 3y = 16</span>Note that the x-terms
would cancel out if only they'd had opposite signs. I can create this
cancellation by multiplying either one of the equations by –1,
and then adding down as usual. It doesn't matter which equation I choose,
as long as I am careful to multiply the –1<span> through the entire equation. (That means both sides of the "equals"
sign!)</span>I'll multiply the second
equation.The "–1<span>R2</span>"
notation over the arrow indicates that I multiplied row 2 by –1.
Now I can solve the equation "<span>–5y = –25</span>" to get <span>y = 5</span>. Back-solving in
the first equation, I get:<span><span>x –
2(5) = –9
x – 10 = –9
x = 1</span>Then the solution
is <span>(x, y) = (1, 5)</span>.</span></span>
A very common temptation
is to write the solution in the form "(first number I found, second
number I found)". Sometimes, though, as in this case, you find the y-value
first and then the x-value
second, and of course in points the x-value
comes first. So just be careful to write the coordinates for your solutions
correctly.
Copyright
© Elizabeth Stapel 2003-2011 All Rights Reserved
<span>Solve the following
system using addition.<span>2x – y<span> = 9
3</span>x + 4y = –14</span>Nothing cancels here, but
I can multiply to create a cancellation. I can multiply the first equation
by 4,
and this will set up the y-terms
to cancel.Solving this, I get that <span>x = 2</span>. I'll use the first
equation for backsolving, because the coefficients are smaller.<span><span>2(2) – y = 9
4 – y = 9
–y = 5
y = –5</span>The solution is <span>(x, y) = (2, –5)</span>.</span></span>
<span>Solve the following
system using addition.
<span>
<span><span>
<span><span><span><span /></span></span></span>
<span><span><span /></span></span>
</span>
<span>
<span /></span></span></span></span><span><span>4x – 3y<span> = 25
–3</span>x + 8y = 10</span>Hmm... nothing cancels.
But I can multiply to create a cancellation. In this case, neither variable
is the obvious choice for cancellation. I can multiply to convert the x-terms
to <span>12x</span>'s
or the y-terms
to <span>24y</span>'s.
Since I'm lazy and 12 is smaller than 24,
I'll multiply to cancel the x-terms.
(I would get the same answer in the end if I set up the y-terms
to cancel. It's not that how I'm doing it is "the right way";
it was just my choice. You could make a different choice, and that would
be just as correct.)I will multiply the first
row by 3 and the second row by 4;
then I'll add down and solve.
Solving, I get that <span>y = 5</span>. Neither equation
looks particularly better than the other for back-solving, so I'll flip
a coin and use the first equation.<span>4x –
3(5) = 25
4x – 15 = 25
4x = 40
x = 10</span>Remembering to put the x-coordinate
first in the solution, I get:<span>(x, y) = (10, 5)</span></span>
Usually when you are solving
"by addition", you will need to create the cancellation. Warning:
The most common mistake is to forget to multiply all the way through the
equation, multiplying on both sides of the "equals" sign. Be
careful of this.
<span>Solve the following
using addition.<span>12x – 13y<span> = 2
–6</span>x + 6.5y = –2</span>I think I'll multiply the
second equation by 2;
this will at least get rid of the decimal place.Oops! This result isn't
true! So this is an inconsistent system (two parallel lines) with no
solution (with no intersection point).no solution</span>
<span>Solve the following
using addition.<span>12x – 3y<span> = 6
4</span>x – y = 2</span>I think it'll be simplest
to cancel off the y-terms,
so I'll multiply the second row by –3.Well, yes, but...? I already
knew that zero equals zero. So this is a dependent system, and, solving
for "<span>y =</span>", the solution
is:<span>y = 4x – 2</span></span>
(Your text may format the
answer as "<span>(s,
4s – 2)</span><span>",
or something like that.)</span>
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