To solve the following linear system by elimination, Brent first multiplied each equation by 10. Explain why he did this step. Complete the solution. 0.3x - 0.5y = 1.2 0.7x - 0.2y = -0.1
HI!!
Hey
multiply by 10 get rid of annoying decimals
\[0.3x - 0.5y = 1.2\\ 0.7x - 0.2y = -0.1\] \[3x - 5y = 12\\ 7x - 2y = -1\]
so are you saying that he multiplied 10 by 0.3....multiplied 10 by 0.5 and multiplied 10 by 1.2?
yeah if u multiply one part by 10, u gotta multiply all by 10
Okay so the answer is: Brent multiplied by 10 to get rid of the decimal numbers
yes
\[1+1+1=3\]\[10+10+10=3\times 10\]
The numbers are still there, I think Brent multiplied by 10 to get rid of the decimal places.
Okay should i add in what misty said before about how ... you'd get 3x -5y = 12
@doc.brown ^
no
\[3x - 5y = 12\\ 7x - 2y = -1\] u still have 2 do more work
okay so is there anything that i should calculate at all?
multiply the first by 2, the second by -5 and add the y terms will go bye bye
Only if you want to follow the instructions, which say complete the solution.
@misty1212 and after doing that I'll find the solution, right?
If i multiply the first equation by 2 and the second by -5 and add the y terms
I'm still a little confused on what I'm supposed to write down
You want to know what x is. One way, is to divide the first one by 3.
0.3 divided by 3= 0.1 0.5 divided by 3= 0.16 1.2 divided by 3=0.4
Ah, sorry, I thought you were starting from\[3x - 5y = 12\\ 7x - 2y = -1\]
oops
3 divided by 3= 1 5 divided by 3=1.6 12 divided by 3=4
I'd do this, it's less work.\[3x - 5y = 12\\\frac{3x}{3} - \frac{5y}{3} = \frac{12}{3}\\\frac{\cancel{3}x}{\cancel{3}} - \frac{5y}{3} = \frac{12}{3}\\x-\frac{5}{3}y=4\\x=4+\frac{5}{3}y\]
thank you what's x and y
I don't know, stick x into the second equation.
did you find it already?
or is that what i'm supposed to do but btw i'm not exactly sure how to solve everything from here. Fractions confuse me.
No they don't, your teacher confused you. @misty1212 was having you make the y's equal and subtract them to get x by itself. I was getting x alone on one side and substituting it in to the other equation to get y by itself. Do you have a preference?
Nah but I'll just follow what you're doing and get x by itself
I'll do whatever it takes to get the correct answer!
so was misty's answer wrong?
They all work, from the first equation I got\[\color{red}{x=4+\frac{5}{3}y}\]Stick that in here\[7\color{red}{x}-2y=-1\]
7(4 + 5/3y) -2y = -1
Distribute the 7. 7(4)+7(5y/3)...
28+ 35y/3
\[28+\frac{35}{3}y-2y=-1\]Now what?
do i bring 28 to the other side?
then it would become -1-28
yup, then add the ys
okay so 37/3 ?
hold on, shouldn't i be subtracting not adding?
then the answer is: 33/3y=-29
@doc.brown ^
Can you imagine a half?
why, what do you mean?
What happens if you add a quarter to a half, what do you get?
3/4
That's right, the bottom number has to be the same and you add up the tops.
\[\frac{35}{3}y-\frac{6}{3}y\]
@doc.brown 29/3
\[\frac{29}{3}y=-29\]
yes
@doc.brown what's the next step
divide both sides but by what
You want the y to be alone, multiply each side by 3/29
1y=-3
Good, so if we know \(y=-3\) What's\[x=4+\frac{5}{3}y\]
x=4+5/3(-3) =-1 so the answer is -1 @doc.brown
Does it work, does \(x=-1\) and \(y=-3\) if\[0.3x - 0.5y = 1.2\]
0.3(-1) - 0.5 (-3) =1.2 Yea it works
@doc.brown Thank you so much for all the help!
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