Question

Solve the following pair of equations by substitution method 0.2z + 0.3y = 1.3 ; 0.4x + 0.5x = 2.3​

Answer 1

Answer:

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Step-by-step explanation:

Let 0.2x+0.3y=1.3       ....(1)

    0.4x+0.5y=2.3       ....(2)

Multiply eqn(1) by 10 we get

2x+3y=13                ....(3)

Multiply eqn(2) by 10 we get

4x+5y=23                ....(4)

Solving equations (3) and (4)

5×(3)⇒10x+15y=65

3×(4)⇒12x+15y=69

Subtracting we get 12x+15y−10x−15y=69−65=4

or 2x=4 or x=2

Substituting x=2 in eqn(3) we get

2x+3y=13⇒2×2+3y=13 or 3y=13−4=9 or y=3

Hence x=2,y=3

Answer 2

Answer:

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Step-by-step explanation:

Let 0.2x+0.3y=1.3 ....(1)

0.4x+0.5y=2.3 ....(2)

Multiply eqn(1) by 10 we get

2x+3y=13 ....(3)

Multiply eqn(2) by 10 we get

4x+5y=23 ....(4)

Solving equations (3) and (4)

5×(3)⇒10x+15y=65

3×(4)⇒12x+15y=69

Subtracting we get 12x+15y−10x−15y=69−65=4

or 2x=4 or x=2

Substituting x=2 in eqn(3) we get

2x+3y=13⇒2×2+3y=13 or 3y=13−4=9 or y=3

Hence x=2,y=3