### Video Transcript

Solve the simultaneous equations π₯ plus four π¦ is equal to 17 and two π₯ plus seven
π¦ is equal to five.

Well, these are simultaneous equations so Iβm going to put these curly braces
here to indicate that theyβre both true at the same time, and Iβm also going to number them. This
is equation one, and this is equation two.

Now the numbering means that I can refer to each equation
individually as Iβm going through and annotating my working out. And looking at both of these
equations, Iβve got π₯ in the first equation and two π₯ in the second equation; Iβve got four π¦ in the
first and seven π¦ in the second. I canβt simply add or subtract one of those equations from the
other to eliminate one of the variables, so what I need to do is to multiply or divide through in
one of the equations to get the same number of π₯s or π¦s in both equations.

Now the easiest way to do this is to double everything in equation one. So doubling everything in equation one, Iβve
got two lots of π₯ are two π₯, and two lots of positive four π¦ are positive eight π¦, and two
lots of 17 are 34, and we can call this equation number three. Now if I look at equations two and
three, Iβve got two π₯ in both of them, and Iβve got seven π¦ in one, eight π¦ in the other, and
then equals five and equals 34.

So if I take one equation away from the other, Iβm going to
eliminate the π₯ variable. But which one should I take away from which? Well if I do equation two
take away equation three, two π₯ minus two π₯ is nothing, seven π¦ take away eight π¦ is gonna
leave me with negative π¦. But if I did equation three take away equation two, eight π¦ minus seven
π¦ will be a positive one π¦, and two π₯ minus two π₯ is still zero. So thatβs what Iβm gonna do.

So equation three take away equation two, two π₯ take away two π₯ is nothing; eight π¦ take away
seven π¦ is just π¦; and 34 take away five is 29. So π¦ is equal to 29. Now I can use that value
for π¦ in one of my earlier equations β one, two, or three β to find out the corresponding value of π₯.
Now the easiest equation to do that with is equation one here.

So substituting π¦ equals 29
into equation one, Iβve got π₯ plus four times 29, four times π¦, is equal to 17; and four times 29 is
116. So π₯ plus 116 is equal to 17. Now if I take away 116 from both sides, Iβm just gonna leave
myself with π₯ on the left-hand side, and 17 take-away 116 is negative 99. So now Iβve got a value
for π₯. Now I can check my answer by substituting in those values Iβve got for π₯ and π¦ into
one of my equations.

Well Iβve already used equation one to work out the value of π₯, so Iβm gonna
use equation two to do my check. So substituting in, π₯ equals negative 99, and π¦ equals 29. Iβve
got two times negative 99 plus seven times 29 is equal to five. Well two times negative 99 is
negative 198, and seven times 29 is 203. So negative 198 plus 203 is equal to five? Well, yes it is!

So that means that the values I used for π₯ and π¦ must be correct. So I can write my answer
out nice and clearly: π₯ equals negative 99, and π¦ equals 29. And I know that my answer is correct
because Iβve checked it.