1.

Solve the following equations without transposing and check your result. (i) x + 5 = 9 (ii) y – 12 = -5 (iii) 3x + 4 = 19 (iv) 9z = 81 (v) 3x + 8 = 5x + 2 (vi) 5y + 10 = 4y – 10

Answer»

i) x + 5 = 9 

x + 5 – 5 9 – 5 (subtract 5 from both sides) 

x = 4 

Check LHS = x + 5 (substituting x = 4) 

= 4 + 5 = 9 

RHS = 9 

∴ L.H.S = R.H.S

ii) y – 12 = – 5 

y – 12 = – 5 

y – 12 + 12= – 5 + 12 (add l2on both sides) 

y = 7 

Check 

LHS = y – 12

= 7 – 12= – 5 

RHS = -5 

∴ L.H.S = R.H.S

iii) 3x+4= 19 

3x + 4 = 19 

3x + 4 – 4 = 19 – 4 (subtract 4 from both sides) 

3x = 15

\(\frac {3x}{3} = \frac {15}{3}\) (Divide both sides by 3)

x = 5 

Check 

LHS

 = 3x + 4 

= 3 x 5 + 4 

= 15 + 4 = 19 

RHS = 19 

∴ L.,H.S = R.H.S

iv) 9z = 81

\(\frac {9z}{9} = \frac {81}{9}\) (Divide both sides by 9)

z = 9 

Check 

LHS = 9z = 9 x 9 = 81 

RHS = 81 

∴ LHS = RHS

v) 3x + 8 = 5x + 2 

3x + 8 = 5x + 2 

3x + 8 – 8 = 5x + 2 – 8 

(Adding -8 on both sides) 

3x = 5x – 6 

3x – 5x = 5x – 6 – 5x 

(Subtract 5x from both sides)

-2x = -6

\(\frac {-2x}{-2} = \frac {-6}{-2}\) (Divide both sides by -2)

Check 

LHS = 3x + 8 = 3(3) + 8 = 9 + 8 = 17 

RHS = 5x + 2 = 5(3) + 2 = 15 + 2 = 17 

∴ LHS = RHS

(vi) 5y + 10 = 4y – 10 

5y + 10 = 4y – 10 

5y + 10 – 1o = 4y – 10 – 10 (Subtract 10 from both sides) 

5y =4y – 20

5y – 4y = 4y – 20 – 4y (Substract ty from both sides) 

y = – 20 

Check 

LHS = 5y + 10 = 5 x( – 20) + 10 = – 100+ 10= – 90

RHS = 4y – 10 = 4 x ( – 20) – 10 = – 80 – 10 = -90

∴ LHS = RHS



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