Question

Solve the following system. (I cant do it)
x+y+z=4
4x+5y=3
y-3z=-10

Answer 1

have you covered solving "system of equations of two variables" yet?

Answer 2

yea I have just been working on this for too long and I cant find an answer

Answer 3

k

Answer 4

gimme a few

Answer 5

\(\large {
\begin{array}{llll}
x+y+z&=4\\
4x+5y&=3\\
y-3z&=-10
\end{array}\implies
\begin{array}{cllll}
x+y+z&=4\\
4x+5y+{\color{brown}{ 0z}}&=3\\
{\color{brown}{ 0x}}+y-3z&=-10
\end{array}
\\ \quad \\
\textit{now let us pick the 1st two only}
\\ \quad \\
\begin{array}{cllcll}
x+y+z&=4&{\color{red}{ \times -4}}\implies &\cancel{ -4x}-4y-4z=-16\\
4x+5y+0z&=3&&\cancel{ 4x}+5y+0z=3\\
\hline\\
&&&\square ?+\square ?+\square ?=\square ?

\end{array}
}\)

what would that give us as resulting equation?

Answer 6

9y-4z=-13??

Answer 7

-4y+5y = 9y? *cough*

Answer 8

but anyhow, the others are ok,
so \(\large \begin{array}{cllcll}
x+y+z&=4&{\color{red}{ \times -4}}\implies &\cancel{ -4x}-4y-4z=-16\\
4x+5y+0z&=3&&\cancel{ 4x}+5y+0z=3\\
\hline\\
&&&0x+y-4z=-13

\end{array}\)

so if we solve there for "y"
what would that give us?

Answer 9

0x+y-4z = -13
or
y-4z = -13

solve for "y" would give us?

Answer 10

it would give us a decimal right?

Answer 11

nope

Answer 12

it'd give us a value, in "z' terms

Answer 13

i have no clue i hate math

Answer 14

hehe
well.... you'd need to solve it like you'd any other linear simplification

say let us use "a" and "b"


if you had a - 4b = 3
solving for "a" would give you?

Answer 15

a - 4b = -13 rather
what would you get for "a"?

Answer 16

hint: add 4b to both sides

Answer 17

a=-13=4b

Answer 18

well... if you add +4b to both sides
you'd get ?

Answer 19

a=13+4b

Answer 20

a=-13+4b

Answer 21

well....yes.. it just that is -13 though

Answer 22

yea sorry

Answer 23

which system was that apart of?

Answer 24

one sec

Answer 25

ohh wel... I simply changed the letters
if a -4b = -13 => a = -13 + 4b

then

y - 4z = -13 => y = -13+4z

Answer 26

But how does that help solve the system, if i now have y equaled to something?

Answer 27

\(\large {
\begin{array}{llll}
(1)x+y+z&=4\\
(2)4x+5y&=3\\
(3){\color{blue}{ y}}-3z&=-10
\end{array}\implies
\begin{array}{cllll}
x+y+z&=4\\
4x+5y+{\color{brown}{ 0z}}&=3\\
{\color{brown}{ 0x}}+y-3z&=-10
\end{array}
\\ \quad \\
y -4z = -13 \implies {\color{blue}{ y}}=\underline{ -13+4z}
\\ \quad \\
thus\qquad (3){\color{blue}{ y}}-3z=-10\implies {\color{blue}{ -13+4z}}-3z=-10
}\)

can you tell what "z" is now?

Answer 28

notice above
since we know now that y =-13+4z
then we could use that in the (3) equation and "substitute" "y" for that
to get "z"

Answer 29

oh i see :D

Answer 30

-13+4z-3z = -10
1z

-13+z = -10
z = -10 + 13

Answer 31

can you seee what "z" is now?

Answer 32

z is 3

Answer 33

yeap z = 3
and y = -13+4z
or
y = -13 + 4(3)

Answer 34

once you have "y" and "z"
you can just plug them in the (1) equation and get "x"

Answer 35

when i check two of them dont work

Answer 36

hmm what values did you get for all 3 variables?

Answer 37

x=-3 y=4 z=3

Answer 38

y = -13+4z
or
y = -13 + 4(3) => y =?

Answer 39

Thank you so much

Answer 40

yw

Answer 41

@jdoe0001 what were your variable answers btw?

Answer 42

ahemm well
y = -13+4z
or
y = -13 + 4(3) => y =-13 + 4*3 => y = -13 + 12
y = -1

we know z = 3 and y = -1
plug them in the 1st equation then
\(\bf x+{\color{brown}{ y}}+{\color{brown}{ z}}=4\implies x+{\color{brown}{ (-1)}}+{\color{brown}{ (3)}}=4\implies x-1+3=4
\\ \quad \\
x+2=4\implies x\cancel{ +2-2}=4-2\implies x=2\)