Question

Solve the following linear equations:
(1)
(I) 9(x-1) = 4(x-3)
(ii) 5m + 2(3m - 7) = 5(3m - 1) +2
(iii)2 m+1/5 = -3
​

Answer 1

Answer:

1

Step-by-step explanation:

1

9x-9 = 4x-12

9x-4x= -12+9

5x = -3

x= -3/5

2

5m + 6m- 14= 15m-5+2

15m -11m = 14-5+2

4m =11

m= 11/4

3

2m = -3-1/5

2m= -16/5

m = -8/5

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Answer 2

(I) 9(x-1) = 4(x-3)

Multiplying the numbers with the numbers in the brackets,

$\twoheadrightarrow \rm 9x - 9 = 4x - 12$

Transposing 4x from R.H.S to L.H.S. Changing it's sign, & Transposing -9 from L.H.S to R.H.S, Changing it's sign,

$\twoheadrightarrow \rm 9x - 4x = -12 + 9$

Performing subtracting and addition,

$\twoheadrightarrow \rm 5x = -3$

Isolating the variable,

$\twoheadrightarrow \rm x = \dfrac{ - 3}{5}$

∴ Hence, The value of x in 9(x-1) = 4(x-3) is -3/5

$\rule{200}2$

(ii) 5m + 2(3m - 7) = 5(3m - 1) +2

Multiplying the numbers with the numbers in the brackets,

${\twoheadrightarrow \rm 5m +6m - 14 = 15m - 5 +2}$

Performing addition and subtraction,

${\twoheadrightarrow \rm 11m - 14 = 15m - 3}$

Transposing -14 to R.H.S & Transposing -3 to L.H.S.

${\twoheadrightarrow \rm 11m - 15m= - 3 + 14}$

Performing subtraction and addition,

${\twoheadrightarrow \rm - 4m= 11}$

Transposing -4 from L.H.S to R.H.S,

${\twoheadrightarrow \rm m= - \dfrac{11}{ 4} }$

∴ Hence, The value of m in 5m + 2(3m - 7) = 5(3m - 1) +2 is - 11/4

$\rule{200}2$

(iii) 2m + 1/5 = -3

Transposing 1/5 from L.H.S to R.H.S

$\twoheadrightarrow \rm 2m = -3 - \dfrac{1}{5}$

Performing subtraction,

$\twoheadrightarrow \rm 2m = -3.2$

Transposing 2 from L.H.S to R.H.S,

$\twoheadrightarrow \rm m = \dfrac{ - 3.2}{2}$

Performing division.

$\twoheadrightarrow \rm m = - 1.6$

∴ Hence, The value of m in 2m + 1/5 = -3 is -1.6.