lol yes.
This is all so cute!
This is an attempt at an explanation for whats going on, not the way that you woul show the solution:
We know that:
2x - 5y = -158 (lets call this equation 1, eq1 for short)
and
x - y = -40 (lets call this equation 2, eq2 for short)
We can add these two equations eq1 and eq2 like so:
(2x - 5y) + (x - y) = (-158) + (-40)
ie:
2x - 5y + x - y = -158 - 40
ie:
3x - 6y = -118
You can understand that: for example you understand a=b and c=d, then a+c=b+d.
Sub in values for a,b,c,d to prove it if your unsure.
In the same way, we can subtract one from the other, eg: a=b and c=d, then it is true that a-c=b-d
Now, this doesnt really get us anywhere, but a usefull thing about equations, is that you can multiply both sides, for example, if you have a=b, then you also have that 2*a=2*b
What we do here, is multiply eq2 by 2:
such that:
2(x - y) = 2(-40)
ie:
2x - 2y = -80
Why?
Well, look at eq1 again: It has '2x'.
eq2 has just 'x', or '1x'.
If you multiply eq2 by 2, you end up with a '2x' term.
And then, if you subtract eq1 from eq2, or the other way around, somewhere within the equation, you end up with 2x - (2x), which results in 0. AND, as the only variables involved are x and y, this means that all the x terms are removed, and your just left with ay=b (a,b constants)
And so y=b/a
So lets continue:
We will multiply eq2 by 2 (ie use '2x - 2y = -80') and subtract that equation from eq1, ie
eq1 - 2*eq2.
So we have:
(2x - 5y) - (2x - 2y) = (-158) - (-80)
ie:
2x - 5y - 2x - (-2y) = -158 - (-80)
ie:
2x - 2x - 5y + 2y = -158 + 80
ie:
0 - 3y = -78
ie:
-3y = -78
ie:
3y = 78
ie:
y = 78/3
ie:
y = 26
NOW, we know what the value of y is! Hooray!
That means we can replace where 'y' is used in the equations, with the value '26'.
So, eq1 becomes:
2x - 5(26) = -158
and eq2 becomes:
x - (26) = -40
We choose one to work with, and we can use it to find out the value of x.
eq2 is easier, as we don't have to * by 5 and divide by 2 etc.
so
x - (26) = -40
ie:
x = -40 + 26
ie:
x = -14
And now we have both values for x and y, and we are done! Hooray*2.
To be safe, you can check by subbing in/substituting both values for x and y into eq1 and e2, and seeing that it all matches up.
..Notice how we went for multiplying eq2 by 2 to get rid of the x, when we could have instead went for 5*eq2, and got rid of the y term.
This would have been harder, as *2 is easier than *5, of course.
Sometimes youll be multiplying both eq1 and eq2. Sometimes you'll be adding instead of subtracting, for example, say we had x + y = 10 (eq1), x - 2y = 4 (eq2). Here, we could multiply eq1 by 2, then add eq2, and because eq2 has (-2y), we'd cancel out the y terms and edn up with 2x = 24 etc.
Have a go at those examples if you'r not already in detention doing them >_>.
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Topic: My simple math homework just got complicated <_< (Read 3627 times)