# Math Help - Tricky linear equations in two variables question

1. ## Tricky linear equations in two variables question

$x/a + y/b = a+ b$

$x/a^2 + y/b^2 = 2$
Ans: $x =a^2, y = b^2$
solve for x and y...I tried solving by elimination and substitution by taking LCM etc but the answer I obtain doesn't match with the answer key.

2. Originally Posted by ice_syncer
$x/a + y/b = a+ b
x/a^2 + y/b^2 = 2$

Ans: $x =a^2, y = b^2$
solve for x and y...I tried solving by elimination and substitution by taking LCM etc but the answer I obtain doesn't match with the answer key.
I'm a bit confused. It looks like you want a solution to the system of two equations for x and y in terms of a and b. But if you plug in your given answer you get two equations in a and b which you can solve for a and b. Are the $x = a^2$ and $y = b^2$ conditions extra equations to put into the system? You can do this (I haven't tried yet) ,but you will not get your given solution.

-Dan

3. Originally Posted by topsquark
I'm a bit confused. It looks like you want a solution to the system of two equations for x and y in terms of a and b. But if you plug in your given answer you get two equations in a and b which you can solve for a and b. Are the $x = a^2$ and $y = b^2$ conditions extra equations to put into the system? You can do this (I haven't tried yet) ,but you will not get your given solution.

-Dan
Well Dan, the answer is x=a^2, y=b^2, well, the solution for both equations is the point of intersection....
for eg:
x-y=0
x+y=0

x=0,y=0 like that...

4. Originally Posted by topsquark
I'm a bit confused. It looks like you want a solution to the system of two equations for x and y in terms of a and b. But if you plug in your given answer you get two equations in a and b which you can solve for a and b. Are the $x = a^2$ and $y = b^2$ conditions extra equations to put into the system? You can do this (I haven't tried yet) ,but you will not get your given solution.

-Dan
Oh ok, I forgot to edit, now check please

5. Originally Posted by ice_syncer

$x/a + y/b = a+ b$

$x/a^2 + y/b^2 = 2$
Ans: $x =a^2, y = b^2$
solve for x and y...I tried solving by elimination and substitution by taking LCM etc but the answer I obtain doesn't match with the answer key.
$bx + ay = ab(a + b)\,\,\,\,\,(1)$

$b^2x + a^2y = 2a^2b^2\,\,\,\,\,\,\, (2)$

Multiply (1) by b and subtract from (2).

$(a^2 - ab)y = 2a^2b^2 - ab^2(a+b) = ab^2(2a - a - b) = ab^2(a-b)$

$a(a - b)y = ab^2(a-b)$

$y = b^2$

Now you can get the other..

6. Originally Posted by Isomorphism
$bx + ay = ab(a + b)\,\,\,\,\,(1)$

$b^2x + a^2y = 2a^2b^2\,\,\,\,\,\,\, (2)$

Multiply (1) by b and subtract from (2).

$(a^2 - ab)y = 2a^2b^2 - ab^2(a+b) = ab^2(2a - a - b) = ab^2(a-b)$

$a(a - b)y = ab^2(a-b)$

$y = b^2$

Now you can get the other..
Too late, I got the answer by the same method some 5 hours ago....
What about the below question ? Are you getting 15?

Father’s age is three times the sum of ages of his two children. After 5 years his age will be twice the sum of age of his two children. Find the age of father.

7. Originally Posted by ice_syncer
Too late, I got the answer by the same method some 5 hours ago....
What about the below question ? Are you getting 15?

Father’s age is three times the sum of ages of his two children. After 5 years his age will be twice the sum of age of his two children. Find the age of father.
Well you can check it on your own instead of asking other people. So do a little backtracking...

Backtracking:
IF fathers age is really 15 then the sum of ages of children must be 5. 5 years from now, his age will be 20 and both the children's age will advance by 5. Thus totally the sum of their ages will be 15 after 5 years. BUT two times 15 is not 20!

Working out:

Let the fathers age be '3x', then the sum of children's age is x. 5 years from now fathers age will be 3x+5, but the sum of children's age will be x+10. Remember that there are two people and both of them advance by 5 years.

Now by given data:

$(3x+5) = 2(x+10) \Rightarrow x = 15$

So fathers age is 3x = 45 years.

Now backtrack this and see if its right!