>> Welcome to the Cypress College Math Review on solving linear equations with fractions.
Objective one.
Solving linear equations involving fractions.
Example. Solve for x in the equation 2/5 times x equals 10.
There are two ways to solve this equation.
In method one, we're going to multiply both sides
of the equation by the denominator, which is 5.
So, on the left hand side, 5 cancels out with 5, and we're left with 2x.
On the right hand side, 10 times 5 is 50.
Now we want to get x by itself by dividing 2 by both sides.
2 divided by 2 is one so we have 1x on the left hand side,
50 divided by 2 is 25, so the answer is x equals 25.
In method 2 we're going to combine the 2 steps that we did in method 1 into 1 step.
That is were going to multiply both sides of the equation by the reciprocal of the fraction.
The fraction is 2/5, so its reciprocal is 5 over 2.
The reason we multiply by the reciprocal is
because 5 halves times 2/5 is 1, so it clears the fraction.
Therefore we have 1x on the left hand side.
On the right hand side we have 10 times 5 over 2.
We can put a 1 underneath the 10, and then multiply across,
10 times 5 is 50, 1 times 2 is 2.
So we have 50 divided by 2, which is 25.
So the answer we got in this method is the same as the answer we got from method 1.
The next example is to solve for x in this equation.
The first thing we need to do is to find the least common denominator.
We see that the denominators are 3, 3, 1, which is underneath the x, and 9.
So the least common denominator of those numbers is 9.
So we multiply the equation by 9.
Now, this is the same as multiplying each term by the LCD.
We have 9 times the first fraction on the left, 9 times the second fraction on the left.
On the right hand side, 9 times the first term, even though this term is not a fraction,
but we need to multiply it by the LCD as well, and then 9 times the second fraction
on the right hand side of the equation.
Now we cross cancel the LCD with the denominators.
And 9 cancels out with 3, on the second term, 9 cancels out with 3, we get 3.
On the right hand side, 9 times x is just 9x, and we have negative 9 cancels
out with 9, and we have negative 1 remained.
Notice that this new equation does not involve fractions anymore.
Now we will distribute.
So 3 times x is 3x, 3 times negative 1 is negative 3, 3 times 2 is 6.
On the right hand side, 9 times x is 9x, negative 1 times 2x is negative 2x,
negative 1 times 3 is negative 3.
Now we combine the like terms.
On the left hand side, negative 3 and 6 are like terms
because they're both constants and they give 3.
On the right hand side, 9x and negative 2x are like terms
because they both contain x. So 9x minus 2x is 7x.
Next we want to gather the terms involving x to the left hand side.
Now you have an option of gathering the terms involved x to the right hand side
if you want to, but it's more convenient to gather on the left hand side
since in the final answer we want x to be on the left hand side.
So here, I want to subtract 7x from both sides of the equation, 3x minus 7x is negative 4x,
then move the 3 down on the left hand side.
On the right hand side, move the negative 3 down, because 7x minus 7x is 0.
Now we want to separate the term that involves x from the constant.
So we subtract 3 from both sides of the equation.
We have negative 4x left on the left hand side, negative 3 minus 3 equal's negative 6.
Finally we can get x by itself by dividing both sides by negative 4.
So I have x is negative 6 over negative 4, which can be reduced to 3 over 2, or 3 halves.
In the next example we want to solve for x in this equation, which contains a fraction on each side
The first thing we need to do is remove the fraction
To do so we will multiply both sides by the L.C.D.
Notice that the L.C.D. for the fractions one tenth and one sixth is 30.
Now we multiply 30 by both sides of the equation. On the left hand side 30 times one tenth is 3.
Notice that we do not multiply 30 times the terms inside the parenthesis.
That's a common mistake that a lot of students make, so you want to be careful on that.
On the right hand side, 30 times one sixth we get 5.
Now the new equation does not involve any more fractions.
So we distribute on the left hand side, 3 times x is 3x, 3 times 1 is 3.
On the right hand side, 5 times 2 is 10, 5 times negative x is negative 5x.
Now we want to move the terms that involve x to the left hand side, so we add 5x to both sides.
On the left, 3x plus 5x is 8x, we move down the 3, and then on the right hand side,
negative 5x plus 5x cancels, so we just have 10 left.
Next we subtract 3 from both sides of the equation, so that we can have x on one side
of the equation and the constant on the other side.
So we have 8x on the left, 10 minus 3 is 7, and then divide by 8 from both sides.
We get x is equal to 7 over 8, this fraction cannot be reduced anymore, so that's the answer.
Pause the video and try these problems.
The next objective is to learn to solve linear equations with answers
as no solution or all real numbers.
Example. Solve for x, 5x minus 1 equals 5 times the quantity of x plus 3.
So we would solve this equation like we normally would.
First we want to distribute on the right hand side, 5 times x is 5x, 5 times 3 is 15.
Now we want to gather the terms that involve x to one side of the equation,
and I choose to gather the x term to the left hand side.
So I subtract 5x from both sides.
It happens that, on the left hand side, 5x minus 5x is 0, so we bring down the negative 1,
on the right hand side, 5x minus 5x is also 0 and we bring down the 15.
Now we see that the new equation does not contain the variable x anymore,
and we get a statement which says negative 1 is equal to 15.
This is a false statement because negative 1 does not equal to 15.
Therefore, the answer is no solution.
This means that we cannot find any value of x such that the left hand side
of the equation will equal to the right hand side.
So, in general, when you solve for the equation, and the variable disappears,
and you get a false statement, like in this case negative 1 equals 15, or 0 equals 5,
negative 6 equals 10, any statement that is false, then your conclusion would be
that the equation has no solution.
The next example is to solve for x in the equation 3x plus 2 times the quantity
of x plus 4 equals 5 times the quantity of x plus 1 plus 3.
We will solve this equation like we normally would by first getting rid of the parenthesis.
So on the left hand side, we want to distribute 2 over.
So we have 2 times x is 2x, 2 times 4 is 8.
On the right hand side, we distribute the 5
over to remove the parenthesis, 5 times x is 5x, 5 times 1 is 5.
And now we combine the like terms on the left hand side,
3x and 2x are like terms and they give 5x.
On the right hand side, 5 plus 3 is 8.
So now we gather x to one side of the equation, so we subtract 5x from both sides.
On the left hand side, 5x minus 5x is 0 and we bring down the 8.
On the right hand side, 5x minus 5x is also 0, and we bring down the 8.
Just like the previous example, the variable x disappears.
And now we get a statement that says 8 is equal to 8.
This is a true statement because 8 is indeed equal to 8.
So, the answer is all real numbers.
Which means that no matter what value of x you plug into the equation,
the left hand side would equal to the right hand side.
So in general, when you solve an equation, and the variable disappears, and you get a statement
that says a number equals a number and it's a true statement, like in this case,
8 is equal to 8, then you would conclude that the answer is all real numbers.
Pause the video and try these problems.