how do u do this problem or just give me the answer. 3x+y=5 3x-3y=21?

Question

i just need help on this

Answer 1

3x+y=5

3x-3y=21

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I like adding equations, never subtracting. So let's change the first equation making it all negative.

-3x + - y = -5

and now add equations...

-3x + - y = -5

3x - 3y = 21


0x - 4 y = 16

Divide by -4 and y = -4
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Plug in your y into either original equation.
3x`+`( y)`=`5
3x`+`(-4)`=`5

Add 4
3x````````= 9

Divide by 3
x```` = `````3

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Again x = 3, y = -4

Answer 2

Oo aargh! You've got your self there a simultaneous equation problem... dont worry-there's a pirate on hand. Your value's of x and y are the same in both equations, we manipulate this.

Take your 1st equation and re-arrange to get a value for y:
3x+y=5
y=5-3x

Then get your 2nd equation, and insert your value for y into it:
3x-3y=21
3x-3(5-3x)=21

This new equation has only x and values in it. By rearranging you can find the true numerical value of x!
First of all lets multiply out those brackets:
(-3 times 5 is minus 15, and -3 times -3x is 9x)
3x-14+9x=21
Rearrange to find x:
12x=36
x=3

Your next step is easy, enter your value of x into your 1st equation to find the numerical value of y:
3x+y=5
3(3)+y=5
y=-4

Then if your bored you can double check and enter your values of x and y into your 2nd equation to see if the equation balances:
3x-3y=21
3(3) -3(-4) =21
9+12=21

Answer 3

You really aren't paying attention in class are you?
Those are 2 seperate problems.They're both easy.I won't even show my work.
The first one is cake..X=1 and Y=2.
3xX=3...add that to Y (which equals 2) and you have 5.
The other problem is a little harder,but I can answer that one too.
X=10 and Y=3....3xX=30...3xY=9....take 3Y(9) from 3x(30) and you get 21.

Answer 4

y=2
i dont no im guessing