Question

eek please someone help me with this elimination problem i have no clue at all
5x-4y=14
3x+3y=3

Answer 1

solve by addition (elimination)

Answer 2

solve for y?

Answer 3

and x

Answer 4

ok 5(0)-4y=14

Answer 5

5x-4(0)=14

Answer 6

What?

Answer 7

and why is there two =s on the second one

Answer 8

is that not what your doing? I'm sorry it looked like what i learned

Answer 9

\[5x-4y=14\]\[ 3x+3y=3\]multiply the first equation all the way across by 3, the second by 4

Answer 10

\[3(5x-4y)=3\times 14\]\[15x-12y=42\] for the first one

Answer 11

5x-4y=14 multiply it with 3
3x+3y=3 multiply with 5
15x-12y = 42
15x+15y= 15
- - -
0x -3y = 27
y = -9
just put this value of y in any equation you will get the value of x

Answer 12

\[4(3x+3y)=4\times 3\]\[12x+12y=12\] for the second one

Answer 13

put them on top of each other and add in a column, get
\[15x-12y=42\]\[12x+12y=12\] add in a column get
\[27x=54\] as the \(y\)'s add up to zero

Answer 14

therefore \(x=\frac{54}{27}=2\) now that you have \(x=2\) you can replace \(x\) by \(2\) in either equation and solve for \(y\)

Answer 15

btw the second equation can be rewritten, by dividing both sides by 3, as
\[x+y=1\] then you could write \(y=1-x\) and replace the \(y\) in the first equation by \(1-x\) to solve
\[5x-4(1-x)=14\] for \(x\) you will get the same answer