so theres this equation and its really confusing me out!

Question

Solve

Y=5x-4y

X+y=12

 

1/3x+y=1/2

X-y=3/4

 

 

Answer 1

Mimi,

 

Let's turn your "out" into an "in", as in becoming clued in on how substitution works.

 

First in general, you need as many independent equations as you have unknown to get a solution.  If you have more equations than unknowns, there's is a "good possibility" that the extra equations add no new information about the relationships between the variables (and are merely redundant, not independent) so they can be ignored.  Or there is the "bad possibility" that the extra equations do add new information of a contradictory nature to make a solution impossible.  You have 2 equations with 2 unknowns in each problem set.

 

Set (1):   y=5x-4y

               x+y=12

Pick an equation that has the simplest solution for x or y in terms of the other unknown variable.  Based on the unity coefficients, pick equation 2 first and arbitrarily solve for x in terms of y:

               x = (12 - y)

Now plug or substitute the expression on the right into the first equation of the pair, everywhere you see an x.  This requires only one substitution, rather than 2 if I had solved for y initially.  So my arithmetic is easier:

               y = 5(12 - y) - 4y = 60 -5y -4y = 60 - 9y,  so 10y = 60  and y = 6.

Knowing y now, solve for x = (12 - y) = 12 - 6 = 6.

 

Note that in a general case, x may appear as a power (xⁿ), or in some function (log 2x, e-^3x2, sin x, etc), then you still substitute your expression for x in terms of y and perform the indicated operation (apply nth power, the log, etc).  So the math could get complicated keeping track of all the details.

 

Set 2:     (1/3)x +  y = (1/2)

                     x  -  y = (3/4)

As an aside given how I have lined up these two equations, if I just add them, the y's drop out.  I can then solve for x, and plug that value into the second equation and get y.

 

But since I have to use the method of substitution here, I'll first solve the second equation for y = x - (3/4).  No matter which variable I choose for substitution the final answer will be the same, just the amount of math manipulation may differ.

 

Substituting for every y in equation 1 you get:   (1/3)x + [x - (3/4)] = (1/2).  Multiply left and right sides of the equality by 12 to clear the denominators:

4x + 12x - 9  =  6,  or  16 x = 15  and x = (15/16)

And  y = x - (3/4)  =  (15/16) - (3/4)  =  (15/16)  -  (12/16)  = (3/16).

 

To check you answers just plug in your solution values into both equations to make sure the equalities hold.

Answer 2

I'm assuming that the x,y,X and Y values are all separate, and that these four equations are all part of the same system. Based on this, the easiest thing to do would probably be cancellation and substitution.

 

If you look at equation (2) X+y=12 and equation (4) X-y=3/4, you can shift them so that they are written as X=12-y and X=3/4 +y. Since X is equal to itself, you can write 12-y=3/4+y.

 

Solving for this, you get: 2y=11.25 (or 11 and 1/4), so y = 5.625 or 45/8

 

Then you plug this y value into (2) or (4) to solve for X, and you can plug y in to (3) as well to solve for x, then once you have x and y you can find Y with (1) easily enough.

 

It would probably be best to keep your answers in fraction form, since that's the form the equations were given in. So be careful with reciprocals and cross-canceling and such! Good luck!