maf help plssssssssssssssssssssssssss
System of Equations by Elimination
16x-4y=28 -8x+3y=-5
You want both numbers to cancel each other out You can choose to remove either variable, which one do you want to remove? x or y?
y
brb rq
You have to add the equations in order to solve for the first variable
After that you'll get your value. You have to plug that value into the other equations in order to solve for the remaining variables
"plug" ?
kk You want to find the lowest common factor between 4 and 3 Should be pretty straight forward 3: 3,6,9,[12],15,18,21 4: 4,8,[12],16,20 Lowest common factor is 12 Multiply each equation by its respective factor to get 12y \(3\times(16x-4y=28)\)= \(48x-12y=84\) \(4\times(-8x+3y=-5)\)= \(-32x+12y=-20\) We have: \(48x-12y=84\) \(-32x+12y=-20\) Now you can add them up Does this make sense?
Ok makes tiny bit sense
mm what confuses you?
e v e r y t h i n g
this part
add ?
Yes add them up \[~~~~~~~48x−12y=84 \\+−32x+12y=−20\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\]
look at dude man, so inspirational.
"different signs you subtract and take the sign of the bigger number" right ?
(4,3) is the answer I assume now
Close y isn't 3 Once you have x, you can just substitute it into either equation Easiest would be to substitute into the second equation (You can do it with the first too) \(-8x+3y=-5\) \(-8\cdot4+3y=-5\) \(-32+3y=-5\) \(3y=27\) \(y=9\)
Jesus,
Hey I tried tho
Thank you, Appreciate your time <3
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