Question

Can Someone Explain this to me?
5x-3y-16=0
4x-5y-5=0
Solution Set:
x| 2 |5 5(2)-3-16 5(5)-3y-16=0
y|-2 |3 10-3y-16 25-36-16=0
3y=16-10 36=16-25
3y=6 3y=9
3 3 y=-2 3 3 y=3

x|5|0 4(5)-5y-5=0 -5y-5=0
y|3|-1 20-5y=0 -5y=5
-5

Answer 1

While it appears to be a system of equations, what are you asked to do and what are you stuck on?

Answer 2

everything i dont know how they get the absolute value of
5x-3y-16=0 which is x| 2 |5
y|-2 |3

Answer 3

I do not believe that is an absolute value, it appears to be a separation for the values.

Answer 4

x| |
y| | this is blank and you have to find the absolute value of this 5x-3y-16=0 to get it.

can you help me understand this?

Answer 5

dd u try what i said??....if u put x=2 in that eqn...what value of y do u get???

Answer 6

You do not have to find the absolute value, if that is what they wanted they would have given you something like |x|=?
You are confusing yourself by worrying about the ABS, don't. As @Sourav16 suggested, just use a plug and chug method.

Answer 7

yea..its a trial and error prob...

Answer 8

can you help me solve this
1.) 2x-3y=2
4x+5y=26?

Answer 9

How would you like to solve it, substitution or elimination?

Answer 10

like the example that have given.

Answer 11

The example you are given is just illustrating how you can take a table of values for x, plug them in, and calculate the value of y. Do you have a list of x and/or y values?

Answer 12

i have no list of x and y values, just that. i have to find x and y also

Answer 13

Well that will be easy then. Take your first equation, 2x-3y=2, pick three (3) values for 'x' and plug them in one at a time, solve for y.

Answer 14

first select a variable...if u select x...keep y unchanged in the equation...then put values of x...starting from 1....ultimately...u'll get a value of x which satisfies the equation...let me solve one of the eqns...
\[2x-3y=2 => 2x-2=3y =>y =2x-3/3\]
now...put x=1...u'll get y=0
put x=2...u'll get y=2/3....but thats fractional....u cant select this value of y...
put x= 4, u get y= 2.....
remember..if u want to do it easily...put the values of x in that last step of the equation i simplified....dat'll be a lot easier and quicker....

Answer 15

Your example equation should read \[y =\frac{2}{3}x-\frac{2}{3}\]

Answer 16

can you guys give me the x here then ill do it from there
x+y-6=0
3x-5y-34=0

Answer 17

getting x is now the only thing i dont know.

Answer 18

You make it up. Pick some values, plug in these and fill in the resulting y.
x | y
------
-2 |
-1 |
0 |
1 |
2 |

Answer 19

2(4)-3y=2
8-3y=2
-3y=2-8
-3y=-6 x|4|
-3 -3 = y|2|

i need another one but is this right?

Answer 20

Yes, that is done perfectly. Excellent work.

Answer 21

another is
2(7)-3y=2
14-3y=2
-36=2-14
-3y=-12
-3 -3 = y=4

Answer 22

Again, correct.

Answer 23

what if y has no value like this?
x+y-6=0
3x-5y-34=0

Answer 24

A variable that stands alone has an invisible 1, so 'y' is the same as '1y' if it is easier for you to use.

Answer 25

4x+5y=26

4(4)+5y=26 4(9)+5y=26
5y=26-16 5y=26-36 x|4|9
5y=10 y=2 5y=10 y=2 y|2|2| am i correct? and the signs also

Answer 26

On your 9/2 values you dropped a -
26-36= -10

Answer 27

x|4|9
so its y|2|-2|

Answer 28

Precisely.

Answer 29

can you help me find the x and y of this? x+y-6=0 i cant seem to get it

Answer 30

It is the same process, you pick a value for x and solve for y, or pick a value for y and solve for x.