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anonymous
 4 years ago
Can someone help me solve the following system of equations. Please explain how.
20a + b = 3
b^2 = 4ac
100a + 10b + c = 10
As you may see I'm trying to find a parabola.
anonymous
 4 years ago
Can someone help me solve the following system of equations. Please explain how. 20a + b = 3 b^2 = 4ac 100a + 10b + c = 10 As you may see I'm trying to find a parabola.

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jhonyy9
 4 years ago
Best ResponseYou've already chosen the best response.1b from the first equation b=320a and substitute in second (320a)^2 =4ac 9 2(3)(20a)+40a^2 =4ac 9+6(20a)+40a^2 =4ac 9120a+40a^2 =4ac

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.0rewrite 20a + b = 3 into the form b = 20a  3 (1) rewrite 100a + 10b + c=10 as c = 10  100a  10b (2) Substitute 1 into 2 c = 10  100a 10(20a  3) c = 100a  20 (3) substitute it into the equation (1) and (3) into b^2 = 4ac then (20a 3)^2 = 4a(100a  20) expand, collect like terms and then solve for a

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The answers are not integers.\[\left\{a=\frac{9}{40},b=\frac{15}{2},c=\frac{125}{2}\right\} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0b=320a Applying in the 2nd eqn (320a)^2=4ac 9+120a+400a^2=4ac Applying in the 3rd eqn 100a30200a+c=10 c=100a+40 applying in 9+120a+400a^2=4ac 9+120a+400a^2=4a(100a+40) 9+120a+400a^2=400a^2+160a 9=40a a=9/40 b=39/2 =15/2 c=125/2
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