Solve the system:] 3x+4y-z=0 x-2y+z=-8 4x+3y-2z=-4
Make z the subject of the first equation z=3x+4y [a] Substitute it in the second x-2y+(3x+4y)=-8 x-2y+3x+4y=-8 4x+2y=-8 4x=-8-2y x=-2-1/2y Substitute x in [a] z=3(-2-1/2y)+4y z=-6-3/2y+4y z=2.5y-6 So now we have x and z written in terms of y. Substitute z and x in the third 4x+3y-2z=-4 4(-2-0.5y)+3y-2(2.5y-6)=-4 -8-2y+3y-5y+12=-4 -4y+4=-4 -4y=-8 y=2 Solve for x and z: x=-2-1/2y x=-2-1=-3 z=2.5y-6 z=2.5(2)-6=-1 Whoa, that was long. I do hope I didn't make a miscalculation. Do check the answers by plugging them into the original equations. Hope it helps.
??? Thanks so much... I will
you're welcome. i hope the ??? mean u understand :)
yes dont worry the questions mark were there b4 you answer... thanks so much
no prob. best of luck =)
I do have a nother question can I ask you?
sure
i hope it's not on systems....can't do that calc again :)
no lol
cool.
The sum of three numbers is 108 . The first number is 8 less than the second . The third number is 2 times the first. What are the numbers?
I'll guide you through this, so you'll do most of the thinking. So, lets let the first number be, say, x. okay?
ok
nice. first: x "The first number is 8 less than the second" so the second can be written as...?
x-8
no Y-8
not quite, but you have the idea. . the first number is 8 less than the second. meaning the second is greater than the first by 8
thus the second is...?
The sum of three numbers is 108 . The first number is 8 less than the second . The third number is 2 times the first. What are the numbers? I'm sure you figured out this one eventually. Let the first number be x. The second will be (x+8) "The first number is 8 less than the second" The third will be 2x. "The third number is 2 times the first" Now, the sum of the three is 108. So we have a nice equation. x+(x+8)+2x=108 Solve for x, then you can calculate the value of the second and third numbers. Best wishes.
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