Download Solve using elimination. When we are using the elimination method

Solve using elimination. When we are using the elimination method to solve a system we want to look
for a variable that are opposites. In this case we have y and minus y.
This is always good because when we add the two equations together they will eliminate. Add the two
equations together we 2 x plus one x gives us 3x. y minus y is 0 so there is no need to write it. Equals 5
plus 3 which is 8. Solve our equation for x by dividing both sides by 3. Gives us x equals eight-thirds.
Find y by using one of the original equations. It doesn’t matter which one as you will get the same
answer for both. Use the one you find easier to work with. I’m choosing the second one - x minus y
equals 3. I’ll substitute our x so I have eight-thirds minus y equals. I want to change my 3 to a fraction
because I know I have to combine it with a fraction. Change three to nine-thirds. Move the eight-thirds
by subtracting from both sides This eliminates our eight-thirds and remember this minus in from of the y
must stay with the y. We have negative y equals nine-thirds minus eight-thirds gives us one-third.
Last step is to take care of the negative. If we divide everything by a negative 1, we get y equals negative
one-third. Our final step is to check our values in the original equations. Always check in both original
equations.
Our first check we’ll substitute into the first equation. We get two times eight-thirds plus negative onethird and we’re checking to see if it is equal to five. Simplify the multiplication we have sixteen-thirds
plus negative one third. Still checking if equal to five. We have like denominators so can add the
numerators, which gives fifteen-thirds. We see this simplifies to five so it does check out.
Also need to check the second equation. Substitute eight-thirds minus y, which is negative one-third
and check if equal to three. When subtracting a negative that changes to addition. We have eightthirds plus one-third. Check if equal to three. Have like denominators. Adding the numerators gives
nine-thirds, which is equal to three.The final answer as an ordered pair is eight-thirds comma negative
one-third. Solve using elimination. Notice we do not have variable that are opposites but we do have
variables that are exactly the same. The y in both equations are both positive.
What we want to do in this case is multiply the second equation by a negative one so we end up with
opposite variable. When we multiply by negative one we need to multiply every term on both sides of
the equal sign. Which gives us negative x minus y equals negative two. I’ve done nothing to the first
equation so I want to copy it exactlythe way it is in the original problem.
I want to add the two equations together. We have 3x minus x gives us 2x. Our y minus y will cancel
out. We have the equal and four minus two is two. Next we’ll divide both sides by two. Gives x equals
one.
Now that we have x, we need to find y. Go back into one of our original equations. I’ll choose the
second because I won’t need to worry about multiplying x by a number. So we’ll use x plus y equals two.
Substitute y for x. One plus y equals two. Subtract one from both sides. Gives y equals one.
Need to check our values. Substitute into our first equation we have three times one plus one and
checking to see if it is equal to four. Three times 1 is three. Three plus one is four. So it checks out in
our first equation. Check in our second equation we substitute one for x plus our one for y and check to
see if it is equal to two. Obviously two does equal two so it does check out. Our answer as an ordered
pair would be one comma one.
Solve using elimination. Here we have two equations and none of the variables are the same or
opposites. So we need to multiply one of the equations by a number so we can eliminate a variable.
Notice in the second equation we have the minus four y. Whenever we have a subtraction we want to
see if the other equation is an addition so we can hopefully avoid multiplying by a negative. In this case,
if we multiply the two y by two, we end up with four y and can eliminate y. So let’s multiply the first
equation by two. When we multiply, we want to multiply every term on both sides of the equal sign.
Which gives us eight x plus four y equal four.
I did nothing to the second equation so I copy it down exactly the way it is in the original problem. Add
the two equations together. Eight x plus three x gives us eleven x. Our y’s eliminate because four y
minus four y is zero. We have equals four plus eighteen gives us twenty-two. Divide both sides by
eleven. x equals two. Now it’s time to find y. Go back to either original equation. I’ll choose the first
one because then I don’t need to deal with dividing by a negative to solve for y. So we’ll use the four x
plus two y equals two. Substitute in x. We have four times two plus two y equals two. Eight plus two y
equals two. Subtract eight from both sides. Two y equals negative six. Divide both sides by two. y
equals negative three.
Final step is to check the answers into both equations. My first equation we substitute the x and y we
get four times two plus two times negative three. We’re checking to see if it’s equal to two. Four times
two is eight. Two times negative three is negative and eight minus six is two. So it checks out.
Checking my second equation. Three times two minus four time negative three. We want to see if that
is equal to eighteen. Three times two is six. Negative four times negative three is positive twelve. Six
plus twelve is indeed equal to eighteen. So it checks out. Our solution as an ordered pair is two comma
negative three.
Solve using elimination. We notice in this system our variables are not the same in both equation nor
are they opposite, or can we multiply one variable to get the variable in the second equation. What we
need to do in this situation is multiply both equations by a different number so we can eliminate a
variable. When the signs for the variable are the same in both equations, it doesn’t matter which
variable you eliminate. I’m going to eliminate the x. In order to eliminate the x, I need to multiply the
first equation by four. So I end up with a twelve x. And the second equation by a three. Now one other
thing is went I multiply I want to make sure my variables are opposites. So one of the equations need
to be negative. Either negative four or negative three. It doesn’t matter which number you make
negative but if you look at the entire equation try to choose a number that will make your variable
positive. In this case we'll eliminate our x and solve for y. So we want to try to make that y positive
when we add together. So a quick check I see I have four time four would give me sixteen. Three time
five gives me fifteen. So in my first equation my y is going to be the larger number. I want that to be
positive so when I add them together I end up with a positive y instead of a negative. So I’m going to
make this four a negative four. Distribute the negative four to every term of the equation on both sides.
We get negative twelve x plus sixteen x because a negative time a negative gives us a positive. Equals
negative forty-four.
For the second equation we distribute to get twelve x minus fifteen y gives us forty-two. We add the
two equations together. Our x’s will cancel out. We have sixteen y minus fifteen y gives us a positive y.
Negative forty-four plus forty-two gives us negative two. We have our y we need to find our x. in one of
our original equations. Again it doesn’t matter which one we choose. I’m choosing the first one only
because the three is a smaller number than four. Otherwise I don’t see much difference between the
two.
We’re going to find our y. Three x minus four y is equal to eleven. Substitute in the negative two for y
give us three x minus four times negative two equals eleven. Multiply. Negative four times negative two
gives us positive eight. Need to move the eight to the opposite side. So subtract eight from both sides.
Gives us three x equals three. Divid by three.x equals one.
Now that I have my xand y, I want to check those value to se if they work in both original equation. So
check in the first equation.Substitute x is one. So we have three times one minus four time our y which
is negative two. We want to see if that is equal to eleven. Do our multiplication we have three.
Negative four times negative two gives us plus eight. Three plus eight is eleven. So it checks out in our
first equation.
Check our second equation. Four times one minus five times negative two equals fourteen. We are
checking to see if that is true. Do our multiplication. We have four plus ten equals fourteen. So it
checks out.
Write our answer as an ordered pair. Our ordered pair is one comma negative two. Notice even
though we found our y first, the y is second in our ordered pair becauseit is always x comma y.