Question

6. Using the elimination method, solve each of the following pairs of simultaneous equations. (C) 5x – 3y - 2 = 0 x+5y-6=0​

Answer 1

Answer:

x + y =5 and 2x –3y = 4

By elimination method

x + y =5 ... (i)

2x –3y = 4 ... (ii)

Multiplying equation (i) by (ii), we get

2x + 2y = 10 ... (iii)

2x –3y = 4 ... (ii)

Subtracting equation (ii) from equation (iii), we get

5y = 6

y = 6/5

Putting the value in equation (i), we get

x = 5 - (6/5) = 19/5

Hence, x = 19/5 and y = 6/5

By substitution methodx + y = 5 ... (i)

Subtracting y both side, we get

x = 5 - y ... (iv)

Putting the value of x in equation (ii) we get

2(5 – y) – 3y = 4

-5y = - 6

y = -6/-5 = 6/5

Putting the value of y in equation (iv) we get

x = 5 – 6/5

x = 19/5

Hence, x = 19/5 and y = 6/5 again

(ii) 3x + 4y = 10 and 2x – 2y = 2

By elimination method

3x + 4y = 10 .... (i)

2x – 2y = 2 ... (ii)

Multiplying equation (ii) by 2, we get

4x – 4y = 4 ... (iii)

3x + 4y = 10 ... (i)

Adding equation (i) and (iii), we get

7x + 0 = 14

Dividing both side by 7, we get

x = 14/7 = 2

Putting in equation (i), we get

3x + 4y = 10

3(2) + 4y = 10

6 + 4y = 10

4y = 10 – 6

4y = 4

y = 4/4 = 1

Hence, answer is x = 2, y = 1

Answer 2

Step-by-step explanation:

5X-3y-2=0 -------→1

X+5y-6=0--------→2

multiply eq 1 with 1

and eq2 with 5 and subtract

5X-3y-2=0

5X+25y-30=0

-28y+28=0

-28y=-28

y=1

substitute y=1 in eq 1

5x-3(1)-2=0

5x-3-2=0

5x=5

X=1

I hope it is useful

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