Solve the following equations with fractions.25y−3=14\frac25y-3=\frac14

y = 65 8 y=\frac{65}{8} y = 8 65

Solve the following equations with fractions. 2 5 y − 3 = 1 4 \frac25y-3=\frac14

Welcome back. For this problem, we'd like to solve the equation two fifths y minus three equals one fourth using the steps we saw in the concept video, starting with step zero, clearing our fractions by multiplying by the LCD. Our fractions are two fifths and one fourth, so when we find the least common denominator, we're really finding the least common multiple of the denominators five and four, which is 20. So we're gonna multiply both sides of the equation by that LCD of 20. That will get us 20 times two fifths y minus three equals 20 times one fourth. Now we just need to distribute that 20 on both sides of the equation. Let's start on the left with 20 times two fifths y. Now 20 times two is 40, so 20 times two fifths is 40 fifths, and we get 40 fifths y. Now 40 divided by five is just eight, so 40 fifths is eight, and we can rewrite this as eight y. Next, we have 20 times minus three, which is minus 60. On the right side of the equation, we have 20 times one fourth, and one fourth of 20 is just five. So that gets us eight y minus 60 equals five. We no longer have any fractions or decimals, so step zero is all set and we move on to step one. But because we don't have anything we need to distribute or any like terms we need to combine, step one is actually also all set, and we can move on to step two. We wanna get all variable terms on one side of the equation and all constants on the other. We can see that our only variable term is on the left side of the equation, this eight y, so let's keep variable terms on the left and move the constants to the right. That means we need to get rid of this minus 60 on the left, which we can do by adding 60 to both sides of the equation. This cancels the minus 60 on the left, leaving us with eight y, and on the right, we get five plus 60, which is 65. We have variables on the left, constants on the right, so step two is all set. For step three, we need to use division to isolate our variable. We need to get rid of this eight here so we can divide both sides of the equation by eight. This is going to cancel the eight on the left side, leaving us with just y. On the right side, we have 65 divided by eight, which is just 65 eighths. We actually can't simplify it any further. It's okay. Sometimes we do get solutions that are fractions. At this point, step three is all set, and we have the solution y equals 65 eighths. I recommend that you do step four, and you plug in 65 eighths back into the original equation here to make sure that it is the solution. But once you're all set, that is the solution to the equation. Great work.

Solve the following equations with fractions. 2 5 y − 3 = 1 4 \frac25y-3=\frac14

y = 65 8 y=\frac{65}{8} y = 8 65

y = − 55 8 y=-\frac{55}{8} y = − 8 55

2. Linear Equations and Inequalities

Solving Linear Equations

Intermediate Algebra

2. Linear Equations and Inequalities

Solving Linear Equations

Struggling with Intermediate Algebra?

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Multiple Choice

Solve the following equations with fractions.
$\frac{2}{5} y - 3 = \frac{1}{4}$

$y=$\frac{65}{8}$$

$x=520$

$y=-$\frac{55}{8}$$

$x=440$

Verified step by step guidance

Start with the given equation: $\frac{2}{5}y - 3 = \frac{1}{4}$.

Isolate the term with $y$ by adding 3 to both sides: $\frac{2}{5}y = \frac{1}{4} + 3$.

Convert the whole number 3 to a fraction with denominator 4 to combine the right side: $3 = \frac{12}{4}$, so $\frac{1}{4} + \frac{12}{4} = \frac{13}{4}$.

Rewrite the equation as $\frac{2}{5}y = \frac{13}{4}$.

To solve for $y$, multiply both sides by the reciprocal of $\frac{2}{5}$, which is $\frac{5}{2}$: $y = \frac{13}{4} \times \frac{5}{2}$.

6:06m

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Struggling with Intermediate Algebra?

Solve the following equations with fractions.25y−3=14\(\frac\)25y-3=\(\frac\)14

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Solve the following equations with fractions.
$\frac{2}{5} y - 3 = \frac{1}{4}$