0

This is for my 7th grader. He wants to know how to use the strategy of "clearing the denominators" to solve this system. Would appreciate your help.

3/x + 1/y = 4 ;

6/x - 2/y = -2

Pearl
  • 69
  • 2
    The second one isn't an equation. Did you mean $\frac 6x-\frac 2y =-2$? – lulu Apr 20 '20 at 23:49
  • 2
    Assuming that's what you meant, then it's easiest to solve this by introducing the variables $X=\frac 1x$ and $Y=\frac 1y$. With that change, this system is now in standard form. – lulu Apr 20 '20 at 23:50
  • 1
    The easiest way is to set u=1/x and v=1/y. Solve for u and v and get x and y. – herb steinberg Apr 20 '20 at 23:52
  • Hi lulu and herb steinberg. I appreciate your replies. He just learned the method to set u=1/x and v=1/y. However, his teacher has asked them to also solve using the strategy for clearing denominators – Pearl Apr 20 '20 at 23:58

2 Answers2

1

A much better way of solving this problem is to let $u=1/x,v=1/y$ so you have $3u+v=4,6u--2v=-2.$ This system is easy to solve for $u$ and $v$. Once you have found $u$ and $v$, $x=1/u,y=1/v.$

P. Lawrence
  • 5,674
1

The strategy recommended by P. Lawrence and others in the comments will certainly work and is probably simpler. However, clearing the denominators requires a different approach.

"Clearing the denominators" means multiplying by some number which will cancel with all the denominators. In this case, multiplying everything by $xy$ will do: for the first equation, we get $$ 3y + x = 4xy $$ and for the second, we get $$ 6y - 2x = -2xy $$ We can now add two copies of the second equation to the first to get $$ 15y - 3x = 0 $$ So we know that $x = 5y$. Now, we plug this back in to one of the original equations and we can solve for $y$. Using the first equation (though the second would work equally well): $$ 3y + 5y = 20y^2 \implies y = 0 \quad\text{or}\quad y = \frac{2}{5} $$ Since we are dividing by $y$ originally, $y$ cannot be $0$, so $y = 2/5$. Finally, now we know that $x = 5(2/5) = 2$. So the solution is $$ (x, y) = \left( 2, \frac{2}{5} \right) $$

Jack Haviland
  • 21