Q. 44

Question

Solve the system. Use any method you wish.

2x2-3y2+1=06x2-7y2+2=0

Step-by-Step Solution

Verified
Answer

The solution of system of equations are 2,-2,2,2,-2,-2,-2,2.

1Step 1. Given information.

Consider the given question,

2x2-3y2+1=0        ...... (i)6x2-7y2+2=0        ....... (ii)

Multiply equation (i) by x2y2,

2x2y2x2-3x2y2y2+x2y2=02y2-3x2+x2y2=0        ...... (iii)

Multiply equation (ii) by x2y2,

6x2y2x2-7x2y2y2+2x2y2=06y2-7x2+2x2y2=0       ...... (iv)

2Step 2. Multiply equation (iii) by 3 .

Multiply equation (iii) by 3,

6y2-9x2+3x2y2=0      ...... (v)

Subtract equation (v) and (iv),

6y2-9x2+3x2y2-6y2-7x2+2x2y2=06y2-9x2+3x2y2-6y2+7x2-2x2y2=02x2=x2y2y=-2,2

3Step 3. Substitute the values of y in equation (iv).

Substitute y=2 in equation (iv),

622-7x2+2x222=012-7x2+4x2=012=3x2x=2,-2

Substitute y=-2 in equation (iv),

6-22-7x2+2x2-22=012-7x2+4x2=012=3x2x=2,-2

Therefore, the solution sets are 2,-2,2,2,-2,-2,-2,2.