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`