Let 2x – y = 3 ——— equation 1
Let x + 5y = 14 ——— equation 2
Making x the subject in eqn 1, = x = y + 3 / 2 ——— eqn 3
• Put eqn 3 in eqn 2
(y + 3 / 2) + 5y = 14
6y = 14 – 3/2
6y = 25/2
y = 25/12
• put y = 25/12 in eqn 3
x = (25/12 + 3/2)
x = 43/12
Answer:
the answer is the last one, you solve this by comparing the lines from each shape. these shapes are congruent, meaning they're the same, so the like AB on the first shape and EF on the second are the same, and so on if that makes any sense. surely someone else could do a better explanation than me tho
Using theorem about secant segments we can write,
AB*AH=AG*AC
AC=4,
CG=6
AG=AC+CG=4+6=10
AH=3
AB= AH+HB=AH+x=3+x
(3+x)*3=10*4
9+3x=40
3x=40-9
3x=31
x=31/3≈10.3
HB≈10.3
EG=HB/2 (as radius and diameter)
EG=10.3/2≈5.2
