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f(x)=2x+c, g(x)=cx+5, fg(x)=6x+d; c and d are constants. Work out the value of d

f(g(x))=2(cx+5)+c fg(x)=2cx+10+c 2cx=6x 2c=6 c=3 10+c=d d=13
IB
Answered by Imogen B. • Maths tutor
6285 Views

Make x the subject of 3y + 2x = 6y - x/2

Multiply whole equation by 2 6y + 4x = 12y - x Take 6y from both sides 4x = 6y - x Add x to both sides 5x = 6y divid both sides by 5 x = 6/5(y)
BP
Answered by Ben P. • Maths tutor
4022 Views

Multiply out: (x - 2)(x + 5)

This can be seen as the same as (x)(x + 5) + (-2)(x + 5). So you are just multiplying the first value in the first brackets with both the values in the second brackets. This is x x + x 5 which equals x^2 + 5...
TS
Answered by Thomas S. • Maths tutor
6466 Views

What are the roots of y=x^2+5x+6 ?

Roots = where line touches x-axis. Set equation equal to 0 as y=0 at the x-axis. Now we need to solve for the x values at which y=0. Quadratic is now 0=x^2+5x+6. Factorise quadratic: 0=(x+3)(x+2). Now you ca...
SW
Answered by Sam W. • Maths tutor
7774 Views

How do you solve a simultaneous equation such as x+2y=10 and 3x+2y=18?

To solve simulatenous equations there are two main methods, substitution and elimintion. The first method requires the principle where if x is equal to a number, say x=2 then we can substitute this in, for e...
KR
Answered by Katie R. • Maths tutor
5538 Views