Question

How do you solve `3/7x+1/2=3` by clearing the fractions?

Answer 1

`x=35/6`

Explanation:

`3/7x+1/2=3`

`3/7x=3-1/2` isolate terms with variable
`3/7x=\color(crimson)(6/2)-1/2` common denominator on right side
`3/7x=5/2` simplify

`3/7x(\color(teal)(2/2))=5/2(\color(teal)(7/7))` multiply to get common denominator overall
`\color(seagreen)(6/14)x=\color(seagreen)(35/14)`

`x=\color(indigo)(35/14*14/6)` divide (multiply by reciprocal for fractions) to get variable
`x=35/6`

Symbolab step-by-step

Answer 2

`color(blue)(x=35/6` or `color(blue)(5 5/6`

Explanation:

`3/7x+1/2=3`

`:.(6x+7)/14=42/14`

multiply both sides by `14`

`:.6x+7=42`

`:.6x=42-7`

`:.6x=35`

`:.color(blue)(x=35/6` or `color(blue)( 5 5/6`

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check:-

`:.3/7(color(blue)(35/6))+1/2=3`

`:.cancel3/cancel7^color(blue)1xxcancel35^color(blue)5/cancel6^color(blue)2+1/2=3`

`:.5/2+1/2=3`

`6/2=3`

`:.color(blue)(3=3`

Answer 3

`x =5 5/6`

Explanation:

When you have an EQUATION with fractions, you can get rid of the denominators immediately by multiplying through by the LCM of the denominators to cancel them.

`3/7x+1/2=3" "(LCD =color(blue)(14))`

`(color(blue)(14)xx3x)/7+(color(blue)(14)xx1)/2=color(blue)(14)xx3" "(larrxx color(blue)(14))`

`(color(blue)(cancel14^2)xx3x)/cancel7+(color(blue)(cancel14^7)xx1)/cancel2=color(blue)(14)xx3" "larr` cancel denominators

`6x+7 =42" "larr` no fractions!

`6x = 35`

`x = 35/6`

`x = 5 5/6`