How do you solve equations with fractions?

The easiest way to solve these type of equations is to get rid of the fractions as quick as possible.

1) Find a common denominator. In this case, it's 10. So put all the numbers over 10.

4y - 1/10 = 3y + 4/5
40y/10 - 1/10 = 30y/10 + 8/10

2) Multiply the entire equation by 10 to get rid of the denominator.

10(40y/10 - 1/10) = (30y/10 + 8/10)
40y - 1 = 30y + 8

3) Re-arrange the equation:
40y - 30y = 8 + 1
10y = 9
y = 9/10

Move fractions to one side and y terms on the other:
4y - 3y = 1/10 + 4/5 (change side, change sign)
1y = 1/10 + 8/10 = 9/10
y = 9/10

Solve For Y Fractions

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Good job Davis and jun. Davis, jake and jun all show a correct step by step answer to the stated problem. However, the answers to exercises to problems at your level are usually in whole numbers. So I suspect that the stated problem is not exactly as it is in the assignment. Does the assignment ask you to solve (9/2) times x + 7 = 25? We will assume this slight change to your question and give the same steps with a little explanation for why you choose each operation. given (9/2) times x + 7 = 25, look for the operation that is LOWEST in the order of operations. The order of operations you've studied, exercised, memorized... are parenthesis, exponents, * /, + and -. What operation associated to x is lowest? Obviously that is the +7. Now do the opposite of that operation to both sides. The opposite of +7 is -7. So the equation becomes (9/2) times x + 7 - 7= 25 - 7 and that reduces to (9/2) times x = 18 Anytime you have had x divided by a number, you have solved for x by multiplying both sides by that number. Actually, when x was divided by an number it was really being multiplied by 1 over that number. I mean x divided by 5 is really x times (1/5). x divided by 7 is really x times (1/7). To solve for x, you multiplied by the inverse. In the above examples, the inverse of 1/5 is 5. So that is why you solved x/5 by multiplying by 5, and x/7 by multiplying by 7. Likewise, you solved 9x by multiplying by 1/9. (same as dividing by 9) Now back to your problem (9/2) x = 18. Again, look for the lowest operation associated to x. That is the multiplication by (9/2). So the opposite of multiplying by 9/2 is multiplying by the inverse 2/9 (9/2) x times (2/9) = 18 times (2/9). That reduces to x = 4. So your methodology for solving for x is as follows. Do whatever algebra is needed to get x on 1 side of the equation. Look for the function associated to x that is lowest in order of operations. Apply the opposite of that function to both sides of the equation. Repeat until x is alone. This methodology will work even when the functions involve roots. (the opposite of sqrt is squaring both sides)

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RE:
How do you solve equations with fractions?
Could you explain how to do this to me? Here's what the problem looks like: 4y - 1/10 = 3y + 4/5

I'm confused...

4y-3y=4/5+1/10
y=(8+1)/10
y=9/10

4y - (1/10) = 3y + (4/5)
4y - 3y = (4/5) + (1/10)
y = (8/10) + (1/10)
y = 9/10 = 0.9


Verify the answer:
4y - (1/10) = 3y + (4/5)
4(9/10) - (1/10) = 3(9/10) + (8/10)
(36/10) - (1/10) = (27/10) + (8/10)
35/10 = 35/10
7/2 = 7/2

y = 8/10 + 1/10
y = 9/10

4y - 1/10 = 3y + 4/5
4y - 3y = 1/10 + 4/5
y = 1/10 + (4 x 2)/(5 x 2)
y = 1/10 + 8/10
y = 9/10 (0.9)

4y - 1/10 = 3y + 4/5

Subtract -3y from both sides

y - 1/10 = 4/5

Convert 4/5 to 8/10 (4/5 = 8/10)

y - 1/10 = 8/10

Add 1/10 to both sides

y = 9/10

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With a fraction, you need to make sure the denominator (bottom number) is equal.