2-, 3-var systems of eqns: 4/5x-2/3y=7/30,0.01x+0.1y=0.03; &

[Rozanne]

1. 4/5x - 2/3y = 7/30
0.01x+0.1y=0.03

2. 2x + 3y - z = 6
3x - y + z = 1
x + y + z = 1

 

[Deleted member 4993]

Re: three and two linear equations, polynimials

1. 4/5x - 2/3y = 7/30
0.01x+0.1y=0.03

2. 2x + 3y - z = 6
3x - y + z = 1
x + y + z = 1

Please show us your work, indicating exactly where you are stuck - so that we know where to begin to help you.

for a review of methods - go to

http://www.purplemath.com/modules/systlin1.htm

 

[fasteddie65]

Re: three and two linear equations, polynimials

1. (4/5)x - (2/3)y = 7/30
0.01x + 0.1y = 0.03

Eliminate the fractions in the first equation, multiplying by 30: 24x - 20y = 7

Eliminate the decimals in the second equation, multiplying by 100: x + 10y = 3

Multiply the second equation by 2: 2x + 20y = 6

Add the third equation and the fifth equation: 26x = 13
x = 1/2

Substitute 1/2 in for x in the modified second equation: 1/2 + 10y = 3
10y = 2 1/2 = 5/2
y = 5/20 = 1/4

x = 1/2 and y = 1/4

 

[fasteddie65]

Re: three and two linear equations, polynimials

BTW it's a polynomial.

2. 2x + 3y - z = 6
3x - y + z = 1
x + y + z = 1

Add the first two equations to eliminate z: 5x + 2y = 7

Add the first and third equation to eliminate z: 3x + 4y = 7

Multiply the first equation by -2: -10x - 4y = -14

Add the last two equations: -7x = -7 or x = 1

Substitute x = 1 into the fourth equation: 5(1) + 2y = 7
5 + 2y = 7
2y = 2
y = 1

Now substitute x = 1 and y = 1 into the third equation: 1 + 1 + z = 1
2 + z = 1
z = -1

x = 1, y = 1, z = -1

 

[Denis]

Re: three and two linear equations, polynimials

BRAVO oh fast one :!:

 

[mmm4444bot]

BRAVO oh fast one

Not really. Fast Eddie needed more than half a dozen steps to solve the second exercise; it can be done in less than half a dozen steps.

Perhaps, Slow Eddie needs more Coca-Cola to get back up to speed.