Can you solve the following 2 simultaneous equations; y=6x-2 and x^2-4x+19=y?

y=6x-2

x2-4x+19

Both equations are equal to Y, hence can be substituted into each other;

6x-2= x^2-4x+19

Now we only have X terms in our equations

Bring all the like X terms together, and make equation equal to 0

  1. 6x-2= x^2-4x+19
  2. 6x= x^2-4x+19+2
  3. 6x= x^2-4x+21
  4. 0= x^2-4x+21-6x
  5. 0= x^2-10x+21

Now we have a simultaneous equations which we can factorise

0=(x-7)(x-3)

We can now solve this as either x-7 must equal 0 or x-3 must equal 0

0=x-7………………………………x=7

Or

0=x-3………………………………x=3

Hence answer is x=7 or 3

MC
Answered by Mohammed C. Maths tutor

6927 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

How find the values of x when x^2+8x+16=0?


Which of these lines are parallel to y=2x+3? Which are perpendicular? Options: 1) y=5x-4, 2) y=-1/3x+3, 3) y=-1/2x-1, 4) y=2x-2/3


Expand and simplify the following expression: x(5x – 2) – 3(x2 – 2x + 7)


Solve the simultaneous equation: (16^x)/(8^y)=1/4 and (4^x)(2^y)=16


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning