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### Course: Algebra (all content)>Unit 2

Lesson 12: Old school equations with Sal

# Linear equations 4

Solving linear equations with variable expressions in the denominators of fractions. Created by Sal Khan.

## Want to join the conversation?

• My problem is with adding or subtracting values once you go through the motions of the first step when you get rid of the (x) in the denominator. EX: -x+2/4x+2=3 my problem is around step two I have issues with the format of setting up the adding or subtracting. Is their a rule for it. Because in this problem you would need to subtract 12x from both sides but their is a -x on the other side already! Is it suppose to be set up like 12x-x or -x-12x?
• -x+2/4x+2=3
-x+2=12x+6
-x-12x=6-2
-13x=4
x=-4/13
isn't that simple...
you have to just follow some rules....
• what if say the equation is 1/2(8x+4)=3/4(4x-6)
How would you solve that. I got a question like this on my grade 9 math quiz and I totally blacked out.
• you would distribute the 1/2 to 8x + 4
4x+2=3/4 (4x-6)
you would then distribute the 3/4 to 4x-6
4x+2=3x-4.5
subtract 3x from both sides
1x+2= - 4.5
then subtract two from both sides
x= - 6.5
• Parent struggling with 8th grade homework, how do you do this problem 12(2k +11) = 12(2k+1)
• Laura,
You probably are working it correctly, but this equation has no solutions.

12(2k +11) = 12(2k+1) First divide both sides by 12
12/12 * (2k +11) = 12/12 * (2k+1) The 12/12 becomes 1 and disappear.
(2k +11) = (2k+1) Now subtract 2k from both sides
2k-2k+11 = 2k-2k+1 The 2k-2k becomes zero
0+11 = 0+1
11=1 This is false. So the equation has no solution. There is no value you can put in for k to solve this equation.

I hope that helps.
• On the exercises for this video, it is hard to know whether or not to write the answer in a fraction or a decimal. Can something be fixed about that?
• yeah man . jus do it in fraction form. youl get it correct
(1 vote)
• Ok, so I was doing this excersise on kahn acedemy, and it was this..
-4x-3/-3x+1 = 5 I got 8/-19, but the answer was 8/11, and i really don't understand why this is so.
• Start by multiplying both sides of the equation by -3x+1. This gives -4x-3 = -15x+5. Add 15x to both sides, giving 11x-3 = 5. Add 3 to both sides, giving 11x=8. Divide both sides by 11, giving x=8/11.

It looks like you subtracted 15 from both sides instead of adding. This gave you -19x instead of 11x.
• Hi Sal, I have a question on one fractional linear equation I just can't seem to solve.

(5x/3) +9= (x/6) +18

It says the answer is 6 but I just can't seem to understand where I'm going wrong with this. Can you please explain? Thanks!
• So, it goes like this.

(5x/3) +9= (x/6) +18 \subtract 9 from both sides
(5x/3)=(x/6)+9 \subtract (x/6) from both sides
(5x/3)-(x/6)=9 \do some reorganization
(5/3)x-(1/6)x=9 \fractions to the common base
(10/6)x-(1/6)x=9
(9/6)x=9 \divide both sides by 9
(1/6)x=1 \multiply both sides by 6
x=6 \done
7(8-x/4)=77
the answer is -12 but i have tried Everything i can't get it?
• 7(8-x/4)=77 Step1=Simplify,divide both sides of the equation by 7 :(8-x/4)=11
Step2=Open the brackets and -8 from both sides.: -x/4=3
Step3=Multiply both sides by 4 to get rid of denominator.: -x=12
Step4=Multiply both sides by -1 to get x=12 Hope this helps. :)
• You combine "like" terms if there are 2 variables....so how would you combine if there were fractions? eg. ( 2 3/4 x)
(1 vote)
• combine all the fractions that have the same variable:)
• can you make a video about how to solve x+y=24 type problems
• At , how does the "(x+1)" and "x+1" cancel out? (the problem on the left side)