MATH 102 A Algebra

Question:

Answer:

Substituting the value y from the 2nd equation onto equation 1

Therefore, the linear equation does not have a solution.

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You can eliminate r by multiplying equations 1 by 5 or equation 2 by 2.

Eliminating s

3.Using cramer’s Rule D=(30 6

The determinant = A= -5.-6. -11.6=96

2 -11). The determinant=30.-11-2.-5=b= –320

Factorize 1/v and it will cancel with the one to the right

Divide by LCM = n2-12n = n48/n2-12n

Cross multiplying

n2-12 n will cancel

So, n=7

Addition of equation 1 to 14 and 2nd to 4 for elimination

S= The price of each strawberry =$12

P=Price per Peacan =8

Addition of 2nd equation to eliminate

S=price per eeach strawberry =$19

P = Price of each New York =$15

To eliminate x, multiply 2nd equation by 15.

$18750 = y=money for govt-insured bonds

x=money from non-insured bonds =$31250

b. Revenue function=600x=500000+200x

Tickets = X

TRUE.g multiply equation 1+y=8 by 2.

The solution will vary from 8 to 16.

FALSE- elimination refers to a method of eliminating the leading equation of the first equation from each other equation. However, it does not replace the latter equation with the sum.

TRUE- A system that consists of equations involving two variables could have solutions.

TRUE- Marginal cost is a variable in a linear cost equation.

TRUE – An “inconsistence system” is a system that consists of linear equations with no solution.

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