The second of two numbers is 7 times the first. Their sum is 72. Find the numbers.
x = first number
7x = second number
x + 7x = 72
simplify and solve equation
the larger of 2 #s is 5 less than twice the smaller. their sum is 43. find the #s
x = smaller number
2x - 5 = larger number
x + (2x - 5) = 43
the ans is 16... x+2x-5=43..... therefor 3x=48 then divide both sides by 3 to solve for x
9 ,63
Well, well, well! Looks like we've got ourselves a math problem! Let's call the first number "X" and the second number "Y". According to our given information, Y is 7 times X, and their sum is 72.
So, we can set up two equations: Y = 7X and X + Y = 72.
Now, let's substitute the value of Y from the first equation into the second equation. We get X + 7X = 72. Combining like terms, we have 8X = 72.
Dividing both sides by 8 gives us X = 9.
Now, if X = 9, we can substitute this value back into the first equation to find Y. Y = 7(9), which gives us Y = 63.
So, the first number is 9, and the second number is 63. Ta-da!
To solve this problem, let's set up equations to represent the given information.
Let's assume the first number is "x." According to the problem, the second number is 7 times the first, which means the second number is 7x.
The problem states that the sum of these two numbers is 72. So, we can set up the equation:
x + 7x = 72
Combining like terms, we have:
8x = 72
To find the value of x, we need to isolate x on one side of the equation. To do so, we can divide both sides of the equation by 8:
8x/8 = 72/8
x = 9
Therefore, the first number is 9.
To find the second number, which is 7 times the first number, we can multiply 9 by 7:
7 * 9 = 63
Therefore, the second number is 63.
In conclusion, the two numbers are 9 and 63, respectively.