## anonymous one year ago how do i solve x = (2y-3)/(y+1) for y?

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1. anonymous

@campbell_st @mathmate @mathway @aaronq @Teddyiswatshecallsme

2. anonymous

@rishavraj @Nnesha @Owlcoffee @LazyBoy @Shalante

3. anonymous

i multiplied y+1 on both sides is that right

4. anonymous

not sure how to isolate y from there tho

5. campbell_st

if you multiply by the denominator you get $x(y + 1) = 2y - 3$

6. anonymous

i got that far @campbell_st

7. campbell_st

then distribute $xy + x = 2y - 3$

8. campbell_st

subtract 2y from both sides $xy - 2y + x = -3$ factor and you get $y(x - 2) + x = -3$

9. campbell_st

next subtract x $y(x -2) = -x -3$ lastly divide both sides by (x - 2)

10. anonymous

so y = (-x-3) over (x-2)? @campbell_st

11. campbell_st

that's what I got

12. Owlcoffee

Whenever we are told to "solve for a variable" we mean doing the necessary operations to isolate it on any side of the "=" sign. So, on the equation: $x = \frac{(2y-3) }{ (y+1) }$ We are asked to solve for "y" so we will do what is necessary to leave "y" on any side of the equality. It is also important to note that this equation is solved for "x", since "x" is isolated on the left side of the equation. "x" itself represents a fraction with denominator "1" so therefore, we can treat this equation as a proportion: $\frac{ x }{ 1 }= \frac{(2y-3) }{ (y+1) }$ So we will cross multiply: $x(y+1)=2y-3$ Applying distributive property: $xy+x=2y-3$ This is the key step, we will transfer all the "y"s to the left side of the equation and obtain: $xy-2y=-x-3$ Now, we will take common factor "y", since it is repeated in both terms of the left side: $y(x-2)=-x-3$ If we divide both sides by ($$x+2$$) we will obtain: $y=\frac{ -x-3 }{ x-2 }$

13. Owlcoffee

(x-2) *

14. anonymous

thank you! i get it now @Owlcoffee

15. Owlcoffee

No problem.

16. mathmate

@Owlcoffee Whenever we divide an expression by another ( x-2, in this case), it is necessary to specify the condition that x-2$$\ne$$0 to be mathematically correct. This is because the results are invalid if we divide by zero.

17. Owlcoffee

That would be true if the relationship between the variables were functional, since we don't know that, we cannot specify any conditions.