Answer:
<h2>The easiest to solve for is x in the first equation</h2>
Step-by-step explanation:
Given the system of equation, x + 4 y = 14. and 3 x + 2 y = 12, to solve for x, we can use the elimination method of solving simultaneous equation. We need to get y first.
x + 4 y = 14............ 1 * 3
3 x + 2 y = 12 ............ 2 * 1
Lets eliminate x first. Multiply equation 1 by 3 and subtract from equation 2.
3x + 12 y = 42.
3 x + 2 y = 12
Taking the diffrence;
12-2y =42 - 12
10y = 30
y = 3
From equation 1, x = 14-4y
x = 14-4(3)
x = 14-12
x = 2
It can be seen that the easiest way to get the value of x is by using the first equation and we are able to do the substitute easily <u>because the variable x has no coefficient in equation 1 compare to equation 2 </u>as such it will be easier to make the substitute for x in the first equation.
Answer:
2 * 2 * 11
Step-by-step explanation:
Find the set of prime numbers that, when multiplied together, give 44.
:)
A1 = 4
a2 = 5a1 = 5 x 4 = 20
a3 = 5a2 = 5 x 20 = 100
a4 = 5a3 = 5 x 100 = 500
a5 = 5a4 = 5 x 500 = 2,500
Tn = ar^(n-1); where a = 4, r = 5
Tn = 4(5)^(n-1) = 4/5 (5)^n
Explicit formular is Tn = 4/5 (5)^n
Recursive formular is