How do you solve `\frac { x - 1} { 3} + \frac { 2x + 1} { 5} = \frac { 3x - 1} { 4}`?

To solve this the first thing to do is make the equation simpler by eliminating the fractions. We can do this by multiplying each side of the equation by a common denominator, in this case `3 * 5 * 4 = 60`. This will eliminate the fractions and keep the equation balanced:

`color(red)(60)(((x-1))/3 + ((2x + 1))/5) = color(red)(60)((3x - 1))/4`

`(cancel60^20 * (x-1)/cancel3) + (cancel60^12 * ((2x + 1))/5) = cancel60^15((3x - 1))/4`

`20(x - 1) + 12(2x + 1) = 15(3x - 1)`

We can now expand the terms within parenthesis, group and combine like terms on each side of the equation:

`20x - 20 + 24x + 12 = 45x - 15`

`20x + 24x - 20 + 12 = 45x - 15`

`(20 + 24)x - 8 = 45x - 15`

`44x - 8 = 45x - 15`

Now we can isolate the `x` terms on one side of the equation and the constants on the other side of the equation:

`44x - 8 color(red)( - 44x + 15) = 45x - 15 color(red)( - 44x + 15)`

`44x - 44x - 8 + 15 = 45x - 44x - 15 + 15`

`0 - 8 + 15 = 45x - 44x - 0`

`-8 + 15 = 45x - 44x`

`7 = (45 - 44)x`

`7 = x`

`x = 7`

`(x-1)/3+(2x+1)/5=(3x-1)/4`

multiply both sides by 3

`x-1+3(2x+1)/5=3(3x-1)/4`

multiply both sides by 5

`5(x-1)+3(2x+1)=15(3x-1)/4`

multiply both sides by 4

`20(x-1)+12(2x+1)=15(3x-1)`

`20x-20+24x+12=45x-15`

`20x+24x-45x=-15+20-12`

`-x=-7`

multiply both sides by -1

`x=7`

substitute x =7

`((7)-1)/3+(2(7)+1)/5=((3)(7)-1)/4`

`2+3=20/4`

`5=5`