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\u2019t be hard for you to figure out that the person is 30 years old.<\/p>\n
This same statement can also be written as y = 2x, where y is the person's age and x is your age.<\/p>\n
Now, let\u2019s say your brother is half your age plus the person's age. This same problem can now be written as y = 2x + \u00bdx. In this equation, we are trying to find your brother's age as it relates to your age.<\/p>\n
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.<\/p>\n
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.<\/p>\n
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.<\/p>\n
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\u00a0below for y in terms of x.<\/p>\n
2y \u2013 6x = 12<\/p>\n
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.<\/p>\n
x\/5 + 1\/y = 3<\/p>\n
y\/3 \u2013 4x\/6 = 5<\/p>\n
Y\/3 \u2013 4x\/6 = 5<\/p>\n
The first thing we want to achieve is to dissolve the equation so it\u2019s no longer in factions. We add the first two equations together<\/p>\n
In this equation, the value of y in terms of x is 2x + 15<\/p>\n
y + 1\/y = x<\/p>\n
The first thing to do is to try and dissolve the fraction. We can achieve this by multiplying both sides by y:\u00a0y2+ 1 = xy.<\/p>\n
This, however, leaves us with a quadratic equation on our hands:\u00a0y2 \u2013 xy + 1 = 0<\/p>\n
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:<\/p>\n
This means that while has two values. Its one of the following:<\/p>\n
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 \u201ca\u201d is not equal to zero is y = (-b + \u00d6(b2 \u2013 4ac)) \/ 2a\u00a0 or\u00a0 y = ( -b - \u00d6(b2 \u2013 4ac)) \/ 2a .<\/p>\n
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.<\/p>\n
2x = 4\/y + y<\/p>\n
Now, using the quadratic equation ay2 + by + c = 0<\/p>\n
Let\u2019s substitute these values in the roots equation<\/p>\n
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.<\/p>\n
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.<\/p>\n