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Chapter 11: Q. 50 (page 774)

Solve the system. Use any method you wish.

x3-2x2+y2+3y-4=0x-2+y2-yx2=0

Short Answer

Expert verified

The solution of system of equations are 2,1.

Step by step solution

01

Step 1. Given information.

Consider the given question,

x3-2x2+y2+3y-4=0......(i)x-2+y2-yx2=0......(ii)

Multiply equation (ii) by x2,

xx2-2x2+y2-yx2x2=0x2x3-2x2+y2-y=0......(iii)

02

Step 2. Substitute the value of y in equation (iii).

Subtract equations (i) and (iii),

x3-2x2+y2+3y-4-x3-2x2+y2-y=04y-4=0y-1=0y=1

Substitute the value of y in equation (iii),

x3-2x2+12-1=0x2x-2=0x=0,2

x=0 cannot be considered as in equation (ii) x is in the denominator.

Therefore, the solution sets are2,1.

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