Pascal M. answered • 03/31/15
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I will do the first one as an example
Use either of the equation and solve it for either x or y. It is always easier to solve for a variable that is alone (i.e. pick an equation containing "x", not "2x" to solve for x.
So solve the first equation for x
x – y = 9 <=> x = 9 + y
Take that value of x and plug it into the second equation wherever you see "x"
2x + 5y = 4 becomes 2(9+y) + 5y = 4, which simplifies to
18 + 2y + 5y = 4... now combine like terms and solve for y
7y = 4 – 18 <=> 7y = –14 <=> y = –14/7 = –2
Now, take your value of y and plug it back into either equation... I'll pick the first one because it's easier.
x – y = 9 becomes x – (–2) = 9... now solve
x + 2 = 9 <=> x = 7
So your answers are: x = 7, y = –2
To double check, plug both x and y in the other equation and check that you get a true equality
2x + 5y =?= 4 becomes
2(7) + 5(–2) =?= 4
14 – 10 = 4
4 = 4... we have a true equality, so the answer is correct.
You can use this method to figure out x and y in the next two problems.
Note that if you don't have any lone "x" or "y", try to solve for the variable that presents itself as multiples of each other... for example in the last problem, you don't have any lone "x" or "y", but you have 3y in the first equation and 6y in the second equation, which is really just 2(3y).... so you can solve the first equation for 3y and plug it wherever you have 3y in the second equation. This helps avoid fractions, but it does not always work.
