\(\frac{1}{2}\times(6-14)-\frac{1}{3}\times(6-27)-3\\ =3-7-2+9-3\\ =0\)
\(\begin{array}{rlllll}&1&0&5&5&0\\-&&9&8&4&5\\-&-&-&-&-&-\\&&&7&0&5\end{array}\)
\(\begin{array}{rll}6(x+4)&=&3(x-2)\\2(x+4)&=&x-2\color{red}\text{(Divide by 3)}\\ 2x+8&=&x-2\color{red}\text{(Get rid of the brackets)}\\ x+8&=&-2\color{red}\text{(Minus x)}\\ x&=&-10\color{red}\text{(Minus 8)}\end{array} \)
8d-3d-12=53
5d-12=53
5d=65
d=13.
Done!! :D
Or am I really worrying so much?
Melody. My silly question.
b = 2A / h
What if h is 0 and the triangle is a straight line? b would be infinity for any values of A?
\(\frac{2}{3}+\frac{1}{2}\\ = \frac{4}{6}+\frac{3}{6}\\ = \frac{4+3}{6}\\ =\frac{7}{6}\\ =1\frac{1}{6}\)
Oops my mistake, 6 x 34 = 204.
y=23b+32023-204q
