Solve by addition or substitution. 4x-3y=24 8x-4y=56 Can someone tell me step by step how to do this I need to know for my test 2marrow.
solve for x in 1st equation. x=(24+3y)/4, then input x on 2nd eq. 8(24+3y)/4-4y=56. then solve for Y and find its value. then input Y on fisrt equation and find X. hope it helps
\[4x-3y=24[1]\] \[8x-4y=56[2]\] \[[1]\times 2\] \[2(4x-3y)=2(24)\] \[8x-6y=48[3]\] \[[2]-[3]\] \[(8x-4y)-(8x-6y)=(56)-(48)\] \[-4y--6y=8\] \[-4y+6y=8\] \[2y=8\] \[y=4\] To find x, sub in the y-value into one the three equations given by me above. (ie. [1], [2] or [3])
what did you do when you got to (8x-4y)-(8x-6y)=(56)-(48) dont understand what u did to get ur answers
I told you in that step I subtracted equation 3 from equation 2.
[3]-[2]
ok got it u canceled out 8x
[2]-[3] sorry.
Yeah. You have to get rid of a variable.
That's the main thing about these. Getting rid of variables until you have one left.
yeah i just didnt get it but now i see u canceled out 8x and it=-4y- -6
Yep.
Remember to find x.
could you help me find x cause im confused
I told you at the end.
That's the most easiest part when you have one variable and you're finding the other.
yeah that means x cud be 3 different numbers.
no.
Try testing it for all three equations. You get the same x-value.
no matter what happens, you get the same x-value...
so how would I multiply 4 by 24?
by using a calculator or getting better at multiplication. If you want to get better at multiplication I suggest you go back to grade 1.
I ment is that what u want me to do for each one multiply it by the 24,56 and 48?
no Where'd you get that from. The numbers in the brackets are just to tell you the number of equations I used. For example, [1] means the first equation, [2] means the second equation and [3] is the 3rd equation. What don't you get?
Ok x=9 thanks.
You're welcome.
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