# Worked solution of problem similar to M1 3X – 5y = 13 4X + 8y = 7 ```Worked solution of problem similar to M1
3X – 5y = 13
4X + 8y = 7
The trick to solve these is you need to add or subtract one entire equation from the other one to
eliminate one of the unknowns. You can eliminate either X or Y first. I am going to chose to
eliminate X first. With my experience I know I will have smaller number s to deal with if I do.
To eliminate X I first recognise that the lowest common multiple of 4 and 3 is 12.
So I’ll multiply the entire first equation by 4 and the entire second equation by 3.
4 x( 3X-5y = 13) is the same as 12X-20y =52
3 x (4X + 8y = 7) is the same as 12X +24y =21
Now we need to decide if we are adding the equations or subtracting them. In this case I can see
that I need to subtract the second one from the first one to eliminate X altogether
12X
– 20y
= 52
-( 12X
+24y
=21 )
12X-12X = 0
First equation
12X
Minus second equation 12X
Yields new equation
12X-12X =0X =0
with no “X”
therefore
-20y-24y =-44y
52-21 =31
-20y
24y
20y-24y =-44y
=52
=21
=52-21 =31
-44y
=31
Thus -44y = 31 (then multiply both sides of equation by -1)
So 44y = -31 (Then divide both sides of the equation by 44)
So y =-31/44 = 0.704545 (exactly or 0.70 if the answer is to an engineering question that might only
requires accuracy of about +/- 1%. In this case none of the numbers in the problem are expressed to
more than 2 significant figures so it is completely fictitious accuracy to express the answer to more
than three significant figures)
Anyway to work out x is we need to substitute the value of y back into one of the original equations.
3x – 5y = 13
4x + 8y = 7
I am choosing the second one because 8 and 44 have a common factor but it really doesn’t matter
So 4X + 8(-31/44) = 7
So 4X + 4(-31/22) = 7 this is a bit of simplification that someone without a calculator might bother to
do!
So 4X + 2(-31/11) = 7 this is a further bit of simplification that someone without a calculator might
do!
So 4X -62/11 = 7 now we need to add 62/11 to both sides to yield
4X -62/11 +62/11 = 7 + 62/11
4X = 7+62/11 now because I don’t like relying on a calculator a writing down really long numbers and
worrying about precision I would express this as follows the trick here is that 7=77/11. I do this so I
can easily add the whole number to the fraction to yield a fraction.
4X = 77/11 + 62/11 =( 77+62)/11 = 139/11 (now we are ready to divide both sides of the equation by
4X/4 = (139/11)/4
X = 139/44
Now we really need the calculator unless we want to practice long division
X = 3.159091 (exactly or I would probably express it as 3.16 if this was a final number that was
needed to an engineering problem when none of the original numbers were given to more than 2
significant figures)
Merit 1 will be achieved by exam so I recommend getting plenty of practice
Here are some for you to solve
M1 (a)
8X - 8Y = 20
2X + 4Y = 4
M1(b)
3X + 5Y = 12
6X – 10Y = 16
If you want more to solve please ask. Any pair of equations like this has a solution. If you can find a
pair of equations like this that has no solution then let me know. The only example I can think of is if
you write two equations for two parallel lines (these will never intersect at any x,y spot on a graph.)
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