Step-by-step explanation:
(x^4)^3=(x^3)^4 , true
=> x^(4×3) = x^(3×4) = x^12
13^4 x 13^7= (13^4)^7, false
13^(4+7) = 13^11
(13^4)^7 = 13^(4×7) = 13^28
y^5 x y^0/y^3=(y^2)^1 , true
y^5 x y^0/y^3 = y^(5+0-3) = y^2
(y^2)^1 = y^(2×1) = y^2
q^0 x q^5/q^2=(q^3)^2/q^3, true
q^0 x q^5/q^2= q^(0+5-2)= q^3
(q^3)^2/q^3 = q^(3×2-3) = q^3
Option A
The solution is 
<em><u>Solution:</u></em>
<em><u>Given system of equations are:</u></em>
3x + 6y = 1 ------ eqn 1
x - 4y = 1 ------ eqn 2
We have to find solution to system of equations
We can use substitution method
From eqn 2,
x = 1 + 4y -------- eqn 3
Substitute eqn 3 in eqn 1
3(1 + 4y) + 6y = 1
3 + 12y + 6y = 1
18y = 1 - 3
18y = -2
Divide both sides by 18

Substitute the above value of y in eqn 3

Thus solution is 
I think it is 60 years old?
Answer:
Step-by-step explanation:
Here go to this website, it will explain everything
https://www.symbolab.com/solver/quadratic-equation-calculator/2x%5E%7B2%7D-4x+5=0