Help please!! Much appreciated! Solve each system by any method of your choice: 2x-5y=-1 4x-5y=-7
some say elimination is easiest
you can exchange rows 4x-5y=-7 add -1*r2 into r1 2x-5y=-1
So then the 5's would cancel out?
i mean -2 times r2 into r1
you could also do that too C but i rather start from the outer left most variable.
4x-5y=-7 add 1*r2 into r1 2x-5y=-1 6x=-8 2x-5y=-1
Oh my gosh, I am so sorry, my computer shut down on me! haha(: Thank you!
6x=-8 2x-5y=-1 from here you know x=-8/6=-4/3 and you can plug that into the bottom equation 2(-4/3)-5y=-1 and solve for y
i think it looks wrong let me do it again.
2x-5y=-1 4x-5y=-7 add -1r2 to r1 -2x = 6 4x-5y=-7 I see now so -2x=6 is x=-3 now plug that into second equations 4x-5y=-7 4*-3-5y=-7
-12-5y=-7 -5y=5 y=-1
Haha(: Thank you so much! Much appreciated!! GBU!! :)
2x - 5y = -1 -->(-1)2x - 5y = - 1 4x - 5y = - 7 --------------- -2x + 5y = 1 (result of multiplying by -1) 4x - 5y = - 7 ----------------add 2x = -6 x = 6/-2 x = -3 now sub -3 in for x in either of the equations 4x - 5y = - 7 4(-3) - 5y = - 7 -12 - 5y = - 7 -12 + 7 = 5y - 5 = 5y -5/5 = y -1 = y check... 2x - 5y = - 1 2(-3) - 5(-1) = - 1 -6 + 5 = - 1 -1 = -1 (correct) ANSWER : x = - 3, y = -1 or (-3,-1)
Wow! haha(: Thanks so much! Such a bug help!! GBU!!
*big :P
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