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1. 3X+3Y/5=`12/5

Y=-4-5X

Substituting the value of y in the 2nd equation onto the first equation

3x+3/5(-4-5x) =12/5

3x-3x=12/5-12/5

Hence the linear equation has no solution

2.-2r+4s-5t=-16

5r+5s+4t=-7

-5r-3s+t

Eliminating r by multiplying equation 1 by 5 and equation 2 by 2 you get

10r+20s-25=-80

10r+10s+8t=-14

30s-17t=-94

2s+5t=6

Eliminating s

-60s-28t=-188

60s+150t=180

t=2.067

s=-2.1685

r=-5.222

3.Using cramer’s rule D=(30 6

2 -6)

Da=(-5 6

-11 -6)

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

Db=(30 -5

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

1.

1. determinant=-2.2- -2.5=6
1. determinant= -1.-5- 8.1=-3
1. solution= 5xdet(-7 -4  - 7xdet(-1 -4  + 1xdet(-1 -7

-6 1                        -6

1 1)

=-5(-3)- 7(-25)-43=-233

1.   -6b-12-13=-2b-12+15

-4b=28

b=-7

1.  1/v2+1/v=7/v2

Factorise 1/v out it will cancel out with the one on the right hand side of the equation

1/v(1/v -7/v)=-1(1/v)

1/v – 7/v=-1

V=6

1. dividing by the  LCM = n2-12n = 8n-48/n2-12n

cross multiplying: (n+1) (n2 -12n)= (8n-48) (n2 -12n)

n2-12 n will cancel out

so n=7

1. 4p+12s=176

14p+6s=184

Eliminating by multiplying equation 1 by 14 and the 2nd by 4

56p+168s=2464

56p=24s=736

S= price of each strawberry=\$12

P=price of each peacan=8

1. mark=3n+1s=64

Ryan=1n+2s=53

Eliminating by multiplying 2nd equation by 3

3n+1s=64

3n+6s=159

S=price of eeach strawberry=\$19

P=price of each new York =\$15

1.  15x/1oo + 7y/100=6000

x+y =50000

multiplying 2nd equation by 15 to eliminate x

0.15x+ 0.07y=6000

0.15x+0.15y=7500

y=money in govt- insured bonds= \$18750

x=money in non-insured bonds=\$31250

14.

a)Cost function= 400x =500000

1. b) Revenue function=600x= (500000+200x)
2. c) break even point

15 .3x=2800-17y

3x+7(444) = 3800

X=23 tickets

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

The solution will change from 8 to 16

1. FALSE- elimination is a method that employs the method of eliminating the leading variable of the first equation from each latter equation but not by replacing the later equation by the sum of itself
2. TRUE- A system of equations of two variables may or may not have solutions
3. TRUE- The marginal cost is a variable m of a linear cost equation
4. TRUE – An inconsistence system is a system of linear equations consisting of equations with no solution.
Cite This Work

My Assignment Help. (2021). Solutions To Linear Equations And Cramer's Rule: Examples And Practice Problems. Retrieved from https://myassignmenthelp.com/free-samples/math102a-algebra/cross-multiplying.html.

"Solutions To Linear Equations And Cramer's Rule: Examples And Practice Problems." My Assignment Help, 2021, https://myassignmenthelp.com/free-samples/math102a-algebra/cross-multiplying.html.

My Assignment Help (2021) Solutions To Linear Equations And Cramer's Rule: Examples And Practice Problems [Online]. Available from: https://myassignmenthelp.com/free-samples/math102a-algebra/cross-multiplying.html
[Accessed 04 August 2024].

My Assignment Help. 'Solutions To Linear Equations And Cramer's Rule: Examples And Practice Problems' (My Assignment Help, 2021) <https://myassignmenthelp.com/free-samples/math102a-algebra/cross-multiplying.html> accessed 04 August 2024.

My Assignment Help. Solutions To Linear Equations And Cramer's Rule: Examples And Practice Problems [Internet]. My Assignment Help. 2021 [cited 04 August 2024]. Available from: https://myassignmenthelp.com/free-samples/math102a-algebra/cross-multiplying.html.

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