Question

How do you solve `\frac { 2a + 4} { 3} - 1= \frac { 7a - 13} { 5}`?

Answer 1

`a=4`

Explanation:

`"multiply ALL terms by the "color(blue)"lowest common multiple"`
`"of 3 and 5 to eliminate fractions"`

`"lowest common multiple of 3 and 5 is 15"`

`cancel(15)^5xx(2a+4)/cancel(3)^1-15=cancel(15)^3xx(7a-13)/cancel(5)^1`

`rArr5(2a+4)-15=3(7a-13)larrcolor(blue)"no fractions"`

`"distribute brackets on both sides of the equation"`

`10a+20-15=21a-39`

`rArr10a+5=21a-39`

`"subtract 21a from both sides"`

`10a-21a+5=cancel(21a)cancel(-21a)-39`

`rArr-11a+5=-39`

`"subtract 5 from both sides"`

`-11acancel(+5)cancel(-5)=-39-5`

`rArr-11a=-44`

`"divide both sides by "-11`

`(cancel(-11) a)/cancel(-11)=(-44)/(-11)`

`rArra=4`

`color(blue)"As a check"`

Substitute this value into the equation and if both sides are equal then it is the solution.

`"left "=(8+4)/3-1=4-1=3`

`"right "=(28-13)/5=15/5=3`

`rArra=4" is the solution"`

Answer 2

a=4

Explanation:

`(2a+4)/3-1=(7a-13)/5`

First - common denominator:

`(2a+4)/3-1/1=(7a-13)/5 | *(3*1*5=15)`
`=> (15(2a+4))/3-((15)1)/1=(15(7a-13))/5`

--

`(15/5=3 , 15/3=5 , 15/1=15)`

--

`=> 5(2a+4)-15=3(7a-13)`
`=> 5*2a+5*4-15=3*7a-3*13`
`=> 10a+20-15=21a-39`
`=> 10a+5=21a-39`

now we take all the numbers with "a" to the right,
and all the numbers without "a" to the left:

`10a+5=21a-39 | (-10a +39)`
`=> 10a-10a+5+39=21a-10a-39+39`
`=> 0a+44=11a-0`
`=> 44=11a`

now I know that 11a=44, but I want to know a=?
`44=11a | ( :11)`
`=> 44/11=(11a)/11`

--

`44/11=4 , 11/11=1`

--

`=> 4=a`

Answer 3

`a=4`

Explanation:

`(2a+4)/3-1=(7a-13)/5`

`(7a-13)/5-(2a+4)/3=-1`

`[3*(7a-13)-5*(2a+4)]/15=-1`

`(21a-39-10a-20)/15=-1`

`11a-59=-15`

`11a=44`

`a=4`