Systems of Linear Equations —
So we are going to solve this
system of linear equations for 𝑥 and 𝑦 by using the method elimination. So there are steps that we need to
follow absolutely every single time to be able to use this method. And the first step we need to do
is to label the equations one and two.
And the reason we do this is to
help us when we’re trying to work out what we’re actually talking about through the
latest steps, so we know exactly we can pinpoint. Now the next thing we need to do
is really check that the coefficients are the same in front of either the 𝑥s or the
𝑦s. So in this case, we can see that
the coefficients aren’t the same in front of the 𝑥 but they are in front of the
𝑦s, so we check. And then if we check and they
aren’t, we’re gonna look at that in the next example. So step two is check the
coefficient are the same.
Step three is if the signs are the
same, you subtract. So we can see that the signs are
the same in front of the 𝑦s, so we’re going to subtract one equation from the
other. And in this case, we can see it’s
just easier as the second equation is larger than the first equation. We’ve got three 𝑥 and we’ve got
nine and seven, so we’re going to do two minus one.
And this is really where our
labelling starts to come in handy, so we’ve got three 𝑥 minus two 𝑥. Well three 𝑥 minus two 𝑥 is just
𝑥. 𝑦 minus 𝑦, they’ll obviously
cancel out. That’s the reason we we’re doing
this subtraction. And then we’ve got nine minus
seven, which gives us two. So we now have the answer that 𝑥
is equal to two. Well step four will be put into
one. So whichever variable we get here,
either the 𝑥 equals something or 𝑦 is equal something or 𝑎 or 𝑏 or whatever
variable we get, we substitute it back into equation one. So we can see two multiplied by
two plus 𝑦 equals seven.
Well two times two is four. And then subtracting four from
both sides, we get seven minus four we know is three, so 𝑦 equals three. Now we’re done there, but it is
possible we may have made a mistake somewhere, so we want to make sure that we’ve
definitely got it right. And the way that we’ll do that is
by substituting in to two, so we’ll say check in two.
So we’re going to substitute our
value of 𝑥 equals two and our value of 𝑦 equal to three into equation two to try
and find that it works. So we are gonna have three
multiplied by two, so that’s six, plus three, that’s the value of 𝑦, and that’s
going to be equal to nine. So then is six plus three equals
nine? Yes, it is. Give yourself two ticks; you know
you’ve done it correct.
We’re gonna follow these same
steps every single time with elimination, so we’re gonna look at another system of
linear equations following the same steps to find the variables 𝑥 and 𝑦 again.
So now we must solve for 𝑥 and 𝑦
with this system of linear equations, and we may be able to see there’s a couple of
things that are different. So first thing we have to do is
label the equations as before.
But then this time, we’re checking
for the same coefficients. Well there aren’t the same
coefficients in front of the 𝑥s and same in front of the 𝑦s. So first of all, we want to see is
it possible to multiply just one equation by a constant to get the same coefficients
in front of the 𝑥s or the 𝑦s. And the answer to that is no. So we’re gonna have to do is
multiply both equations by a constant. Well fi- we could multiply both
equations by six and four, respectively, but that’s a larger number. So we wanna pick the smaller
numbers always. So we’re going to have to multiply
both the first equation by three, because that’s the coefficient in front of the 𝑦
for equation two, and then the second equation by two, because that’s the
coefficient in front of the 𝑦 for equation one.
So we’re doing these
multiplication to give us the same coefficients in front of the 𝑦s. So by doing each of them, we’ll
label three multiplied by equation one. We’re gonna label that equation
three. And then two multiplied by all of
equation two, we’re gonna label that equation four.
So three now be really, really
careful that you make sure to do multiplying by each term, because this is where
people usually end up losing marks in the answers, by not doing the multiplication
properly. So we’re doing three multiplied by
six 𝑥, which gives us eighteen 𝑥; then three multiplied by two 𝑦 gives us six 𝑦;
and then fifteen multiplied by three is forty-five.
So now moving on to equation
four. We’re gonna multiply each term by
two. So first, two multiplied by four
𝑥 is eight 𝑥; then two multiplied by three 𝑦 is six 𝑦; then two multiplied by
negative three is negative six.
So now we finally completed step
two. We have checked that the
coefficients are now the same in front of the 𝑦s in fact because we’ve got six 𝑦
in both equation three and equation four. So we can finally go onto step
three, which is if the signs are the same you subtract. Well the signs are the same so we
are going to subtract. If it weren’t the same, then we’d
have to add. So we can see that three is larger
than four, so we’ll do equation three take away equation four.
So we’ve got eighteen 𝑥 take away
eight 𝑥 gives us ten 𝑥. Of course the 𝑦s will cancel out
because we have six 𝑦 minus six 𝑦, which is zero 𝑦. And then we’ll have forty-five
minus minus six, be careful, so we add six. So it’s forty-five add six, which
is fifty-one. And that gives us an 𝑥-value of
𝑥 equals fifty-one divided by ten, so five point one.
Then onto step four, put into
one. So we look back to equation one
and we put our 𝑥-value into it. So we’ve got six multiplied by
five point one, which is thirty point six. Then we will add two 𝑦, and that
is equal to fifteen. So then subtracting thirty point
six from both sides, we get two 𝑦 equals negative fifteen point six. And then dividing that by two to
get rid of the two 𝑦, you get that 𝑦 is equal to negative seven point eight.
Hey so moment of truth, we can now
check whether this actually works. So we’re gonna check again in two,
so we’ve got four multiplied by five point one. We’re adding three multiplied by
negative seven point eight, and that should be equal to negative three.
So four multiplied by five point
one is twenty point four and then three multiplied by negative seven point eight is
negative twenty-three point four. So twenty point four minus
twenty-three point four is equal to negative three, so we can give ourselves two
ticks because we’ve got the 𝑥 and the 𝑦 correct. Although they seem like quite
nasty numbers, actually they work for our system of equations.