How do I solve this system of equations using elimination or substitution? 3x+4y=16 -4x-3y=-19 I keep getting fractions using both methods, and I feel like I'm doing them wrong.
Choose one equation and isolate one variable; this equation will be considered the first equation. (Use the method detailed in earlier units to do this.) Substitute the solution from step 1 into the second equation and solve for the variable in the equation. Using the value found in step 2, substitute it into the first equation and solve for the second variable. Substitute the values for both variables into both equations to show they are correct.
So y=-3/4x+4, I isolated the variable. Then since that equation equals y, I plug in this equation into the y of the other system? Like -4x-3(-3/4x+4)=-19?
give me a sec.
Ok. Show me how you did it. Draw it out.
the answers are not fractions...just so you know
So for the first equation, I subtracted the 3x, so 4y=-3x+16. Divide 4, so y=-3/4x+4. Then I plugged that in to the second equation for y, so -4x-3(-3/4x+4)=-19.
They're not fractions?! Ugh.
do you know how to do elimination ?
Or Substitution?
or, I should say, which way is easier for you
Oh wait, elimination worked, I guess I messed up somewhere. I got (4,1)? But I still need to solve with substitution, which is just confusing.
3x + 4y = 16 3x = -4y + 16 x = -4/3y + 16/3 now we sub that into the other equation -4x - 3y = -19 -4(-4/3y + 16/3) - 3y = -19 -- distribute through the parenthesis 16/3y - 64/3 - 3y = -19 -- multiply entire equation by 3(to get rid of fractions) 16y - 64 - 9y = - 57 -- combine like terms 7y - 64 = -57 7y = -57 + 64 7y = 7 y = 1 now sub 1 in for y in either equation to find x 3x + 4y = 16 3x + 4(1) = 16 3x + 4 = 16 3x = 16 - 4 3x = 12 x = 4 solution is : (4,1)
do you have any questions ? Do you see where you messed up ?
I do, I didn't try to isolate x and it was easier that way, thanks!
no problem :)
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