If the sum of an integer and 6 times the next consecutive integer is 61, the the value of lesser integer is 7
Consider the first odd integer as x
Then the next consecutive odd integer = x+2
The 6 times the second integer= 6(x+2)
= 6x+12
Sum of an integer and 6 times the next consecutive odd integer is 61
Then the equation will be
x + 6x+12 = 61
Add the like terms in the equation
(1+6)x + 12 = 61
7x +12 = 61
Move 12 to the right hand side of the equation
7x = 61-12
7x = 49
x = 49/7
x = 7
The second number is
x+2 = 7+2
= 9
Hence, if the sum of an integer and 6 times the next consecutive integer is 61, the the value of lesser integer is 7
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First step find a common multiple, e.g. 45 and subtract
9x + 5y = 28 * 5
5x + 9y = 56 * 9
45x + 25y = 140
-(45x + 81y = 504)
-56y = -364
56y = 364
y = 6.5
Substitute into any equation
9x + 5(6.5) = 28
9x + 32.5 = 28
9x = -4.5
x = -0.5
x = -0.5 , y = 6.5
Answer:-5
Step-by-step explanation:
Hey there!
To figure out the decimal of 7/33, just divide the numerator by the denominator. The numerator is the top part of the fraction. The denominator is the bottom part of the fraction.
So, 7 ÷ 33 = 0.212121.....
We can write this as 0.21 with a bar on top of 21. This shows that 21 is recurring.
Thank you! :D
Answer:
9/16
Step-by-step explanation:
