Exercises 27–40 contain linear equations with constants in denomi...

Question

Exercises 27–40 contain linear equations with constants in denominators. Solve each equation.3x/5 = 2x/3 + 1

Accepted Answer

Welcome back. I'm so glad you're here. We have this equation four Y divided by seven equals why divided by three plus five. And we are to solve four. Why? Alright, how do we do this with these fractions? Well we want to take a look at the fractions and figure out what our common denominator would be. So with seven and three. Well that's just going to be 21. And now we want to take each term and multiply it by our lowest common denominator. So we're going to multiply all three of these terms by 21. So the work here gets easier because we can reduce our factors in the numerator and the denominator. So when we've got 21 divided by seven, that's going to be a 3/1 and 21 divided by three. That's now going to be a 7/1. So reduced this reads three times four. Why Equals seven times y plus five times still 21. Well three times four Y is 12, Y equals seven times Y. Is seven, Y plus five times 21 is 105. Let's get our wise on the same side. Let's subtract seven Y from both sides. So 12 Y minus seven Y. Is five. Y equals the seven wise, cancel out on the right and it equals 105 to get that Y by itself. Let's divide both sides by five. And hey look that's familiar. Why equals 105 divided by five. That is from back here. 21. So why equals 21. We look at our answer choices and this matches with answer choice D. Well done, we'll catch on the next one.

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Exercises 27–40 contain linear equations with constants in denominators. Solve each equation.3x/5 = 2x/3 + 1

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