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Solving For Y in Terms of X Using Fractions

By Kathleen Knowles, 30 Sep 2020

The first step to solving any mathematical equation is to understand the question. For instance, if someone says to you, "my age is twice yours," and you know that you are fifteen years old, then it won’t be hard for you to figure out that the person is 30 years old.

This same statement can also be written as y = 2x, where y is the person's age and x is your age.

Now, let’s say your brother is half your age plus the person's age. This same problem can now be written as y = 2x + ½x. In this equation, we are trying to find your brother's age as it relates to your age.

In this situation, we're assuming that your age is an unknown variable x. It is problems like this that introduce the concepts of solving for one variable in terms of another.

In this section, we'll be solving for Y in terms of X using fractions and the different methods that are involved. There is no one way to solve for Y in Terms of X Using Fractions. You'll get better as you practice and start realizing your own techniques.

Let's Solve for Y

We're going to start with simple examples that solve x in terms of y. Then eventually we'll move on to more complex ones.

Once again, to solve y in terms of x is to find that value of y but not necessarily as a constant but in the form of x. Solve the equations below for y in terms of x.

Example 1

2y – 6x = 12

Solution

  • 2y = 12 + 6x (divide both sides by 2)
  • Y = 6 + 3x

This means that we've found the value of y, but not just as a constant. The value of y we found is dependent on the value of x. If x is 1, then y is 9. If x is 2, then y will be 12, etc.

Example 2

x/5 + 1/y = 3

Solution

  • x/5 + 1/y = 3 (lets add the first two fractions with x and y)
  • (xy + 5 )/ 5y = 3(multiply both sides by 5y)
  • Xy + 5 = 15y (subtract xy from both sides)
  • 5 = 15y –xy(factor out y)
  • 5 = y(15 - x) (divide both sides by (15 - x))
  • 5/ (15 - x) = y
  • Y = 5/ (15 - x)

Example 3

y/3 – 4x/6 = 5

Solution

Y/3 – 4x/6 = 5

The first thing we want to achieve is to dissolve the equation so it’s no longer in factions. We add the first two equations together

  • y/3 – 4x/6 (their lowest common multiple of their denominator is 6, so the equation becomes)
  • (2y – 4x)/ 6 = 5 (cross multiply to dissolve fractions)
  • 2y – 4x = 30 (rearrange to get y)
  • 2y = 4x + 30 (divide both sides by 2)
  • Y = 2x + 15

In this equation, the value of y in terms of x is 2x + 15

Example 4

y + 1/y = x

Solution

The first thing to do is to try and dissolve the fraction. We can achieve this by multiplying both sides by y: y2+ 1 = xy.

This, however, leaves us with a quadratic equation on our hands: y2 – xy + 1 = 0

If you have dealt with quadratic equations before, then you'll be familiar with the rule for ay2+ by + c = 0 where a is not equal to zero, then the equation can also be written as:

  • a(y2 + bx/a) = - c  (divide both sides by a)
  • y2 + by/a = -c/a  (in order to square the left hand side, we add  (b/2a)2 to both sides )
  • y2 + by/a +(b/2a)2 = - c/a + (b/2a)2
  • (y + b/2a)2 = -c/a + b2/4a2 (taking the square root of both sides)
  • y + b/2a = +Ö (-c/a + b2/4a2 ) or
  • X + b/2a = -Ö (-c/a + b2/4a2 ) (now, simplifying for the value of  y, the equation becomes )
  • y =- b/2a ± Ö (-c/a + b2/4a2 )
  • y = ( -b + Ö(b2 – 4ac)) / 2a  or
  • y = ( -b - Ö(b2 – 4ac)) / 2a
  • in the equation Y2 – xy + 1 = 0
  • a = 1,
  • b = -x
  • c = 1

This means that while has two values. Its one of the following:

  • y = ( x + Ö(x2 – 4)) / 2 or
  • y = ( x - Ö(x2 – 4)) / 2

This may appear a bit complex at first, especially the part where we had to find the root of the quadratic equation, but it gets simpler. In the next example, we will just skip the whole steps since we already know that the root of any quadratic equation ay2+ by + c = 0 where “a” is not equal to zero is y = (-b + Ö(b2 – 4ac)) / 2a  or  y = ( -b - Ö(b2 – 4ac)) / 2a .

All we would be doing is find the value of a, b and c. Then we can substitute them in the roots equation to find y.

Example 5

2x = 4/y + y

Solution

  • 2x = 4/y + y (to remove the fraction, multiply both sides by y)
  • 2xy = 4 + y2 (this can also be rearranged to be)
  • y2 - 2xy + 4 = 0

Now, using the quadratic equation ay2 + by + c = 0

  • a = 1
  • b = -2x
  • c = 4

Let’s substitute these values in the roots equation

  • y = ( -b + Ö(b2 – 4ac)) / 2a  or
  • y = ( -b - Ö(b2 – 4ac)) / 2a
  • y = ( -(-2x) + Ö((-2x)2 – 4*1*4)) / 2*1
  • y = ( 2x + Ö(4x2 – 16)) / 2 or
  • y = ( 2x - Ö(4x2 – 16)) / 2

Wrapping up

As you can see from the examples above, there is no one direct way to solve for y in terms of x using fractions. Based on the values you have, you could be solving a linear algebraic equation, simultaneous equations, or a quadratic equation.

Whichever one you end up with is based on the values that you have in the question. First, try and simplify the fractions. This lets you know what kind of equation you'll be solving.

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