Linear Equations

Exercise 2.3 Part 2

Solve the following equations and check your results

Question 6: `8x+4=3(x-1)+7`

Solution: Given `8x+4=3(x-1)+7`

By removing brackets from RHS we get:

`8x+4=3x-3+7`

Or, `8x+4=3x+4`

By transposing 3x to LHS and 4 to RHS we get:

`8x-3x=4-4`

Or, `5x=0`

After dividing both sides by 5 we get:

`(5x)/(5)=0/5`

Or, `x=0`

CHECK: Given equation is `8x+4=3x+4`

By substituting the value of x in RHS we get:

`8x+4=3xx0+4`

Or, `8x+4=0+4=4`

By transposing 4 to RHS we get:

`8x=4-4`

Or, `8x=0`

By dividing both sides by 8 we get:

`(8x)/(8)=0/8`

Or, `x=0` proved


Question 7: `x=4/5(x+10)`

Solution: Given `x=4/5(x+10)`

After removing the brackets we get:

`x=(4x)/(5)+(40)/(5)`

Or, `x=(4x)/(5)+8`

By transposing `(4x)/(5)` to LHS we get:

`x-(4x)/(5)=8`

Or, `(5x-4x)/(5)=8`

Or, `x/5=8`

By multiplying both sides by 5 we get:

`x/5xx5=8xx5`

Or, `x=40`

CHECK: Given equation is: `x=4/5(x+10)`

By substituting the value of x in RHS we get:

`x=4/5(40+10)`

Or, `x=4/5xx50`

Or, `x=4xx10=40` proved


Question 8: `(2x)/(3)+1=(7x)/(15)+3`

Solution: Given `(2x)/(3)+1=(7x)/(15)+3`

By transposing `(7x)/(15)` to LHS and 1 to RHS we get:

`(2x)/(3)-(7x)/(15)=3-1`

Or, `(10x-7x)/(15)=2`

Or, `(3x)/(15)=2`

Or, `x/5=2`

By multiplying both sides by 5 we get:

`x/5xx5=2xx5`

Or, `x=10`

CHECK: Given equation is `(2x)/(3)+1=(7x)/(15)+3`

By substituting the value of x in RHS we get:

`(2x)/(3)+1=(7xx10)/(15)+3`

Or, `(2x)/(3)+1=(14)/(3)+3`

By transposing 1 to RHS we get:

`(2x)/(3)=(14)/(3)+3-1`

Or, `(2x)/(3)=(14)/(3)+2`

Or, `(2x)/(3)=(14+6)/(3)`

Or, `(2x)/(3)=20/3`

By multiplying both sides by 3 we get:

`(2x)/(3)xx3=20/3xx3`

Or, `2x=20`

By dividing both sides by 2 we get:

`(2x)/(2)=20/2`

Or, `x=10` proved

Question 9: `2y+5/3=(26)/(3)-y`

Solution: Given `2y+5/3=(26)/(3)-y`

By transposing –y to LHS and 5/3 to RHS we get:

`2y+y=(26)/(3)-5/3`

Or, `3y=(26-5)/(3)`

Or, `3y=(21)/(3)=7`

By dividing both sides by 3 we get:

`(3y)/(3)=7/3`

Or, `y=7/3`

CHECK: Given equation is `2y+5/3=(26)/(3)-y`

By substituting the value of y in RHS we get:

`2y+5/3=(26)/(3)-7/3`

Or, `2y+5/3=(26-7)/(3)`

Or, `2y+5/3=(19)/(3)`

By transposing 5/3 to RHS we get:

`2y=(19)/(3)-5/3`

Or, `2y=(19-5)/(3)=(14)/(3)`

By dividing both sides by 2 we get:

`(2y)/(2)=(14)/(3)÷2`

Or, `y=7/3` proved

Question 10: `3m=5m-8/5`

Solution: Given `3m=5m-8/5`

By transposing 5m to LHS we get:

`3m-5m=-8/5`

Or, `-2m=-8/5`

Or, `2m=8/5`

By dividing both sides by 2 we get:

`(2m)/(2)=8/5÷2`

Or, `m=4/5`

CHECK: Given equation is `3m=5m-8/5`

By substituting the value of m in RHS we get:

`3m=5xx4/5-8/5`

Or, `3m=4-8/5`

Or, `3m=(20-8)/(5)=(12)/(5)`

By dividing both sides by 3 we get:

`(3m)/(3)=(12)/(5)÷3`

Or, `m=4/5`

CHECK: Given equation is `3m=5m-8/5`

By substituting the value of m in RHS we get:

`3m=5xx4/5-8/5`

Or, `3m=4-8/5`

Or, `3m=(20-8)/(5)`

Or, `3m=(12)/(5)`

By dividing both sides by 3 we get:

`(3m)/(3)=(12)/(5)÷3`

Or, `m=4/5` proved



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