Respuesta :

mathmate

This is basically a system of 3 linear equations with variables x, y and c.


x+y = 6...................................(1)

5x/3 + y = c .........................(2)

2y = c-4x .............................(3)


(2)-(1)

5x/3-x +y-y = c-6 => 2x/3 = c-6 => x = (3/2)(c-6)=3c/2-9 ..............(4)

(3) - 2(2)

2y-2y - 10x/3 = c -4x -2c

=> 2x/3 = -c

=> x = -(3/2)c.............................................................................................(5)


Finally, equate (4) & (5) to get

3c/2-9 = -3c/2

=>

3c = 9

=> c=3

jcherry99

I think, but I wouldn't want to bet serious money on it, that solving for c last is the way to go. Equation 2 and equation 3 both have one as the numerical coefficient for c, so you could equate the rest of the equation with each other.


Step one

Isolate c on Equation 3

Add 4x to both sides.

2y + 4x = c


Step 2

Isolate c in equation two.

It is isolated. Equate the left hand sides of each equation

2y + 4x = (5/3)x + y Multiply by 3


Step 3

Solve for ax + by

6y + 12x = 5x + 3y Subtract 3y from both sides.

3y + 12x = 5x Subtract 5x from by sides.

3y + 7x = 0


Step 4

Use equation 1

3y + 7x = 0

y + x = 6 Multiply this equation by 3


3y + 7x = 0

3y + 3x = 18 Subtract

4x = - 18

x = - 4.5


x + y = 6

-4.5 + y = 6 Add 4.5 to both sides.

y = 10.5


So far we have

x = - 4.5

y = 10.5

Now we have to solve for c


In case I run out of time, c = 3


4x + 2y = c

4*(-4.5) + 2(10.5) = c

-18 + 21= c

c = 3 <<<<<< answer.


I cannot tidy this up. I'm being timed. The answers are correct however.



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