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4z(3z)^2 + 10y - (-3y + 10y) x 2 = 15z - 5y && (((21y + 3y)/2 - 15y) * 2) + 7y + 21z^2 = 7

2006-06-28 03:32:18 · 6 answers · asked by mvolosen 2 in Science & Mathematics Mathematics

4(3z)^2 + 10y - (-3y + 10y) x 2 = 15z - 5y && (((21y + 3y)/2 - 15y) * 2) + 7y + 21z^2 = 7

(mistake was fixed)

2006-06-28 03:37:33 · update #1

(4z(3z))^2 + 10y - (-3y + 10y) x 2 = 15z - 5y && (((21y + 3y)/2 - 15y) * 2) + 7y + 21z^2 = 7

(Sorry, another detail added to make it easier - and I do not know the answer)

2006-06-28 03:48:39 · update #2

(4(3z))^2 + 10y - (-3y + 10y) x 2 = 15z - 5y && (((21y + 3y)/2 - 15y) * 2) + 7y + 21z^2 = 7

Equation 1 should simplify to -12z^2 + 15z = y

2006-06-28 03:53:39 · update #3

6 answers

Jack's simplification is correct.

Equation 1 simplifies to y = -144z^2 + 15z

Equation 2 simplifies to y = -21z^2 + 7

Combining and simplifying we get 123z^2 - 15z + 7 = 0. You can then use the quadratic formula to get the two possible values of z (both will be imaginary) and then plug these into equation 1 or equation 2 to obtain the possible values for y.

Equation 1 will not simplify as the author wrote since (4(3z))^2 = *(12z)^2 = (12)^2 * z^2 = 144z^2.

2006-06-28 06:47:17 · answer #1 · answered by Anonymous · 1 0

I get it down to: 15Z^2 - 15Z + 7 = 0 , but I can't find the roots. Are you sure the question is correct?

Based on your latest revision, equation 1 can be simplified to:
Y = 15Z - 144Z^2
Equation 2: Y = 7 - 21Z^2
Therefore: 123Z^2 - 15Z + 7 = 0
I still can't solve it.

2006-06-28 11:47:25 · answer #2 · answered by Jack 2 · 0 0

after simplifying we get y=-36z^3+15z . hence we have this equation 36z^3-21z^2-15z+7=0 solve it to find z and then you will get the value of y

2006-06-28 10:50:35 · answer #3 · answered by s topology 1 · 0 0

I never could do these kinds of math problems! I always felt that numbers were for math and letters were for writing. I just couldn't combine the two.

2006-06-28 10:35:59 · answer #4 · answered by grahamma 6 · 0 0

i think there is something wrong with the question...coz when i try to find the roots for the solved equation iam not getting the roots

2006-06-28 11:03:56 · answer #5 · answered by lillyqwe2004 3 · 0 0

do you know the answer?

2006-06-28 10:35:46 · answer #6 · answered by pearls3212 4 · 0 0

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