Posted: June 25th, 2022

MATH 102 A Algebra

Question:

Answer:

Substituting y’s value in the 2nd equation to the 1st equation

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The linear equation is therefore not solvable.

Multiply equation 1 by 5, equation 2 by 2, and subtract r to eliminate r.

Eliminating s

3.Using the cramer’s rule D=(30 6)

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The determinant is a= -5.-6 – -11.6=96

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

Factorise 1/v to cancel out the one on the right side

Divide by the LCM = N2-12n = 8n48/n2-12n

Cross multiplying: (n+1), (n2-12n)= (8n48) (n2-12n)

n2-12 and n will cancel each other

So n=7

Multiply equation 1 by 14, and equation 2 by 4 to eliminate

S= Price of each strawberry =$12

P=Price per peacan=8

Multiplying the 2nd equation by 3 will eliminate

S=price for eeach strawberry=$19

P = Price of each New York = $15

Multiply 2nd equation with 15 to get rid of x

$18750 = y=money in government-insured bonds

x = money in non-insured bond=$31250

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

Tickets X = 23

TRUE-e.g multiply equations x+y=8 and 2

The solution will be different from 8-16

FALSE- Elimination is a method which eliminates the leading variable from the first equation of each equation, but not by replacing it with the sum of the equations.

TRUE – A system of equations with two variables can or may not provide solutions

TRUE – The marginal cost of an equation with a linear cost is a variable m.

TRUE – An Inconsistence System is a set of linear equations that contains equations without a solution.

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