How do you solve the simultaneous equations #y = 3x^2- 10# and # 13x - y = 14#?

1 Answer
Jul 26, 2015

Answer:

Substitute the expression for #y# from the first equation into the second to yield a quadratic in #x#. Solve for #x# and hence find #y#.

#(x, y) = (4, 38)# or #(1/3, -29/3)#

Explanation:

Substitute the expression for #y# from the first equation into the second to get:

#14 = 13x-(3x^2-10) = 13x-3x^2+10#

Subtract the right hand side from the left to get:

#3x^2-13x+4 = 0#

Use a version of the AC Method to factorize this:

Let #A=3#, #B=13#, #C=4#

Find a pair of factors of #AC = 3*4 = 12# whose sum is #B=13#.

The pair #B1=12#, #B2=1# works.

Then for each of the pairs #(A, B1)# and #(A, B2)# divide by the HCF (highest common factor) to find a pair of coefficients of a factor (choosing suitable signs):

#(A, B1) = (3, 12) -> (1, 4) -> (x-4)#
#(A, B2) = (3, 1) -> (3, 1) -> (3x-1)#

So:

#0 = 3x^2-13x+4 = (x-4)(3x-1)#

which has roots #x = 4# and #x=1/3#

If #x = 4# then #y = 3x^2-10 = 48-10 = 38#

If #x = 1/3# then #y = 3x^2-10 = 3*1/9-10 = 1/3-10 = 1/3-30/3 = -29/3#