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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

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Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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In Exercises 1-8, solve the system of equations.

{ 3 y 2 + 6 y = 0 x y + x + 2 = 0

To determine

To calculate: The solutions of system of equations 3y2+6y=0 and xy+x+2=0.

Explanation

Given Information:

The system of equations are:

3y2+6y=0xy+x+2=0

Formula used:

Pair of linear equation (linear) in two variables

a1x+b1y=c1a2x+b2y=c2

A system of linear equation can be solved using the method of elimination. To solve the system of equation using elimination follow the following procedure.

Step 1: Arrange both equations such that the like variables and constants are one above the other.

Step 2: Decide which variable to eliminate and with proper choice of multiplication, arrange so that the coefficients of that variable are same or opposite in sign for another.

Step 3: Add the equations if they are of opposite sign and subtract the equations if they are same sign leaving one equation with one variable.

Step 4: Solve for the remaining one variable.

Step 5: Substitute the value found in the above step in any equation involving both variables and solve for the other variable.

Calculation:

Consider the given equations

3y2+6y=0 …… (1)

And

xy+x+2=0 …… (2)

Consider the equation (1) and solve it for y,

3y2+6y=0y(3y+6)=0

Thus,

y=0

And

3y+6=0y=63=2

Now substitute y=0 in the equation (2),

x(0)+x+2=0x+2=0x=2

Now substitute y=2 in equation (2),

x(2)+x+2=02x+x+2=0x+2=0x=2

Thus, x=2,y=2 and x=2,y=0

Now, check x=2,y=2 and x<

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