Solve for x and y on the following systems of equations?

Question

9x/5 + 5y/4 = 47/10
2x/9 +3y/8 = 5/36

another one
x/4 +3y/7 = -2/21
2x/3 + 3y/2 = -7/36

I am pulling my hair out with these!!

Answer 1

9x/5 + 5y/4 = 47/10
2x/9 +3y/8 = 5/36

36x+25y=94
16x+27y=10

y=(94-36x)/25
400x+2538-972x=250

y=-2
x=4

for another one use the same method!

Answer 2

Here's a really fast and easy way to solve systems of equations. It requires that you have a graphing calculator (or some matrix software like MATLAB). I have a TI-83 Plus.

Write out the equations like this:

(9/5)x + (5/4)y = (47/10)
(2/9)x + (3/8)y = (5/36)

Now plug the coefficients into a 2 x 3 matrix, like so:
[ (9/5) (5/4) (47/10) ]
[ (2/9) (3/8) (5/36) ]

(on my calculator you can access matrix operations by pressing 2ND X^-1, MATRX is written in yellow above the button)

Once you have entered the matrix into, say, matrix [A], scroll over to the MATH menu within the MATRX menu and find an operation called:
rref(

Select that, then select the matrix name. Your main screen should look something like this:
rref( [A]

Hit Enter, and you should get something that looks like this:
[ [ 1 0 4 ]
[ 0 1 -2] ]

The first number at the end represents x, the second number represents y, so:

x = 4
y = -2

This also works for more than 2 variables. When you use rref( , you should always get an answer of the form:

[ 1 0 0 0 a
0 1 0 0 b
0 0 1 0 c
0 0 0 1 d ]

Just make sure in your original equations all the x's are lined up, all the y's are lined up, etc. It's hard to explain why this works unless you've taken linear algebra.

For the second problem I get:

[ (1/4) (3/7) (-2/21) ] ~ [ 1 0 (-2/3) ]
[ (2/3) (3/2) (-7/36) ] [ 0 1 (1/6) ]

x = (-2/3)
y = (1/6)

There is a little bit of a learning curve involved, but once you get it, it's so much easier than working these out the other way.

Hope that helps.

Answer 3

To get rid of fractions, multiply each
equation by the least common multiple.

In the equation, 9x/5 + 5y/4 = 47/10, the LCM is 20,
because that is the lowest number that 5, 4 and 10
will each divide into exactly.

You could multiply through by a higher number, say 40 or 60,
but then you would have larger numbers to deal with, as well
as trying to reduce them by cancelling common factors. That's
why the least common multiple is used, not higher multiples.

So, multiplying through by 20 gives : 36x + 25y = 94.

With 2x/9 + 3y/8 = 5/36, the LCM is 72.
So, multiplying through by 72 gives : 16x + 27y = 10.

Now there are a few options for these 2 equations.

1) You could multiply the first one by 27 and the second one
by 25, to eliminate y, but this is getting into big numbers.

2) Or multipling the first one by 16 and the second one
by 36 will eliminate x, but these are still big numbers.

3) But notice that the LCM of 36 and 16 is 144.
That is: 36 * 4 = 144 and 16 * 9 = 144.
That means you could multiply the first equation by 4 and the
second one by 9 to eliminate x. These smaller numbers are
much better to work with.

4) But there's an even easier way.
Just subtract the second one from the first one.
Doing this gives : 20x - 2y = 84
Now divide out the common factor, which is 2.
10x - y = 42
Rearrange to : y = 10x - 42
Now plug this value of y into the second equation,
because that's the simplest one to work with.
This gives : 16x + 27(10x - 42) = 10
Notice that we can divide through by 2 to make the
calculations even simpler. Then we get :
8x + 27(5x - 21) = 5
Now multiply out : 8x + 135x - 567 = 5
or, 143x = 572. Therefore, x = 4.
Substitute this into y = 10x - 42.
We get : y = 10*4 - 42 = -2

Same with the second set of equations.

x/4 +3y/7 = -2/21
LCM of 4, 7 and 21 is 2*2*7*3 = 84.
Multiplying by 84 gives : 21x + 36y = -8.

2x/3 + 3y/2 = -7/36
LCM of 3, 2 and 36 is 36.
Multiplying by 36 gives : 24x + 54y = -7.

Subtract first equation from second one this time.
3x + 18y = 1
We could then find either x or y, but as you can
see, this might complicate matters.
What if we just rearrange to find 18y = 1 - 3x.
Now we can easily substitute this into the first equation
to get : 21x +2(1 - 3x) = -8
or, 21x + 2 - 6x = -8
or, 15x = -10, therefore, x = -2/3.
Substitute this back into the simplest equation to get :
18y = 1 - 3(-2/3) = 3. Therefore, y = 1/6.

This is a different way of solving such equations than
you've probably been taught, but it can come in handy
sometimes to reduce calculations.
It may in fact, have been easier to eliminate y in the
second set of equations by multiplying one by 3 and
the other by 2, after seeing that 36 and 54 are both
simple multiples of 18.
Try some more examples using both methods and
see what you are comfortable with.

Answer 4

9x/5 + 5y/4 = 47/10
2x/9 + 3y/8 = 5/36

X=4,y=-2

x/4 + 3y/7 = -2/21
2x/3 + 3y/2 = -7/36

x= -2/3,y= 1/6

Answer 5

come to a call proper right here structures of equations for x and y a million). y = 3 + 3x y = 18 - 2x 3x - y = -3 2x + y = 18 5x = 15 x = 3 y = 12 2). 18x + 6y = one hundred and twenty 4x + 3y = 40 8x + 6y = 80 10x = 40 x = 4 y = 8

Answer 6

9x/5 + 5y/4 = 47/10 multiply by 20
36x+25y=94
2x/9 +3y/8 = 5/36 multiply by 72
16x+27y=10 multiply by 9
36x+25y=94 multiply by -4
144x+243y=90
-144x-100y=-376 add 2 equations
143y=-286 divide by 143
y=-2 substitute in
16x+27y=10
16x-54=10
16x=64
x=4
solution is (4,-2)

another one
x/4 +3y/7 = -2/21 multiply by 84
21x+36y=-8 multiply by 3
2x/3 + 3y/2 = -7/36 multiply by36
24x+54y=-7 multiply by -2
63x+108y=-24
-48x-108y=14 add
15x=-10 divide by 15
x=-2/3 substitute in
21x+36y=-8
-14+36y=-8 add 14
36y=6 divide
y=1/6
solution is (-2/3,1/6)