I would combine then solve.
L = liz's age now
J = Jack's age now
L-1 and J-1 are their ages one year ago
L-1=3(J-1)
L-1=3J-3
L=3J-3+1
L=3J-2 (to be used later)
L+2 and J+2 are their ages in two years
L+2=2(J+2)
replace L by 3J-2
3J-2+2=2(J+2)
3J=2J+4
3J-2J=4
J=4 is Jack's age now
L-1=3(J-1)
L-1=3(4-1)
L-1=3*3
L-1=9
L=9+1
L=10 is Liz's age now
Answer:
-sqrt(5) ≤ x ≤ sqrt(5)
Step-by-step explanation:
x^2-5≤0
Add 5 to each side
x^2-5+5≤0+5
x^2 ≤5
Take the square root of each side, remembering to flip the inequality for the negative sign. Since this is less than we use and in between
sqrt(x^2) ≤ sqrt(5) and sqrt(x^2) ≥ -sqrt(5)
x ≤ sqrt(5) and x ≥- sqrt(5)
-sqrt(5) ≤ x ≤ sqrt(5)
Match bases
if a^x=b^x where x=x then a=b
hmmm
1/8=8^-1=(2^3)^-1=2^-3
so
2^n=2^-3
n=-3
X6 – 9x3 + 8 = 0
let u = x^3
u^2 - 9u + 8 = 0
(u-8)(u-1) = 0
u = 1,8
back substitute
1 = x^3
8 = x^3
<span>x = 1 , 2</span>
