Question

HELP ME SOMEONE PLEASE :c

Answer 1

Solve the system of linear equations below.

2x + 3y = 2
x + 6y = 4

Answer 2

Multiply the second equation by 2.

Answer 3

x = 4, y = -2

Answer 4

Where did that come from?

Answer 5

No.

Answer 6

x = 2, y = - 2/3

Answer 7

meoowww

Answer 8

\[\Huge{\color{red}{\rightarrow \boxed

{\mathbb{\text{Hello!!}{}}}{\color

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Answer 9

You could use substitution method. You could subtract the 2y and substitute that for x in the first equatipn

Answer 10

YOUR flutterKEN ANNOYING Kuoministers

Answer 11

elimination easier? and FASTER

Answer 12

LOL!!!!!

Answer 13

x = 0, y = 2/3

Answer 14

that felt mean :D

Answer 15

\[2x + 3y = 2 [1]\]
\[x+6y=4[2]\]
\[2 \times [2]\]
\[2(x + 6y)= 2(4)\]
\[(2x + 12y)=8[3]\]
\[[3]-[1]\]
\[(2x+12y)-(2x+3y)=8-2\]
\[2x-2x+12y-3y=6\]
x's cancel out.
\[9y=6\]
\[y=\frac{ 2 }{ 3 }\]

Answer 16

azteck is like the best teacher ever... bow down

Answer 17

Solve the system of linear equations below.

y = -2x + 19
y = x + 7 n this one ?

Answer 18

good work @Azteck

Answer 19

@mariisol please post that separately

Answer 20

You can find x yourself. @mariisol I feel that you need to work on your basic arithmetic because it seems you can't add/divide/multiply/subtract efficiently. Mathematics is a step by step process and practice.

Answer 21

i AGREE WITH AZTECK

Answer 22

Yeah. Keep doing practice problems by yourself. You need to be really good with algebra with a single variables before you move on to complex systems of equations. You are going to have a hard time in linear algebra if you can't add, subtract, multiply or divide lol...

Answer 23

Lol @ linear algebra.