Question 1035244
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Solve the following equation for x and y:

(2)^2+x +(3)^3+y=275 
(2)^3+x +(3)^2+y=145
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<pre>
Since you do not use parentheses properly, there are several ways to read your system.
Your writing is ambiguous. To avoid ambiguity, use parentheses.

Is THIS modification what you want?

{{{2^(2+x) +  3^(3+y)}}} = {{{275}}},     (1)
{{{2^(3+x) +  3^(2+y)}}} = {{{145}}}.     (2)


Below is the solution of this system.


Since {{{2^(2+x)}}} = {{{4*2^x}}}, {{{3^(3+y)}}} = {{{27*3^x}}}, {{{2^(3+x)}}} = {{{8*2^x}}}, {{{3^(2+y)}}} = {{{9*3^x}}}, you can rewrite the system (1), (2) in the form

{{{4*2^x}}} + {{{27*3^y}}} = {{{275}}},      (1')
{{{8*2^x}}} + {{{9*3^y}}} = {{{145}}}.       (2')

This system is still non-linear.
The way to solve it is to introduce new variables u = {{{2^x}}} and v = {{{3^y}}}.
Then the system (1'), (2') takes the form

4u + 27v = 275,         (3)
8u +  9v = 145.         (4)

The system (3), (4) is just a linear, and we can easily solve it.
I will use the Elimination method. Multiply equation (3) by 2 (both sides). You will get

8u + 54v = 550,         (3')
8u +  9v = 145.         (4')

Now, distract (3') from (4'). You will get

54v - 9v = 550 - 145  --->  45v = 405  --->  v = {{{405/45}}} = 9.

Now recall that v = {{{3^y}}},  or  9 = {{{3^y}}}.  Hence,  y = 2,  and one unknown is just found.

Next, from (3)  u = {{{(275 - 27v)/4}}} = {{{(275-27*9)/4}}} = {{{32/4}}} = 8.

Again, recall that  u = {{{2^x}}},  or 8 = {{{2^x}}}.  Hence,  x = 3.

The problem is solved.

<U>Answer</U>.  x = 3,  y = 2.

You can check that the solution is correct.
</pre>