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Chapter 11: Q. 43 (page 794)

In problems 39-43, Solve each system of equations

x2-3x+y2+y=-2x2-xy+y+1=0

Short Answer

Expert verified

Solutions of the system of equationsx2-3x+y2+y=-2x2-xy+y+1=0are(1,0)&(1,-1)

Step by step solution

01

Step 1. Given data 

The given system of equation is

x2-3x+y2+y=-2x2-xy+y+1=0

02

Step 2. Rearrange system of equation

multiply the equation x2-xy+y+1=0by y

role="math" localid="1646953222095" x2-xy+y+1y=0yx2-x+y2+y=0

so a new system of equation is

role="math" localid="1646953230474" x2-3x+y2+y=-2x2-x+y2+y=0

03

Step 3. determine the values of variable

Determine the difference between both equations of a system of equation

x2-3x+y2+y-x2-x+y2+y=-2-0-2x=-2x=1

04

Step 4. Solution of the system of equations  

Substitute x=1in equationx2-3x+y2+y=-2

12-3(1)+y2+y=-2y2+y=0y(y+1)=0y=0,-1

So solutions of the system of equations are(1,0)&(1,-1)

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