the sum of 4 times one number and 3 times a second number is 64. If the sum of the two numbers in 19, find the two new numbers
From the problem statement we derive a system of equations.
4a + 3b = 64
a + b = 19
There are a variety of ways to solve this, but I'll use substitution.
Isolate a variable.
a = 19 - b
... And substitute that into the other equation.
4(19 - b) + 3b = 64
76 - 4b + 3b = 64
-b = -12
b = 12
Plug that into the isolated equation.
a = 19 - 12
a = 7
So our answer is (a, b) = (7, 12).
Let's assume the first number is x and the second number is y.
According to the given information,
4x + 3y = 64 .......(equation 1)
x + y = 19 .........(equation 2)
We can solve this system of equations using the method of substitution or elimination.
Let's solve it using the method of substitution:
From equation 2, we can express x in terms of y:
x = 19 - y
Now substitute x in equation 1:
4(19 - y) + 3y = 64
76 - 4y + 3y = 64
76 - y = 64
-y = 64 - 76
-y = -12
Divide both sides by -1 to solve for y:
y = -12 / -1
y = 12
Now substitute y back in equation 2 to find x:
x + 12 = 19
x = 19 - 12
x = 7
So, the two numbers are x = 7 and y = 12.
To solve this problem, we can set up a system of equations.
Let's call the first number x, and the second number y.
From the given information, we can write two equations:
Equation 1: 4x + 3y = 64 (the sum of 4 times the first number and 3 times the second number is 64)
Equation 2: x + y = 19 (the sum of the two numbers is 19)
To find the values of x and y, we can solve these equations simultaneously:
Step 1: Rearrange Equation 2 to solve for x:
x = 19 - y
Step 2: Substitute the value of x from Step 1 into Equation 1:
4(19 - y) + 3y = 64
Step 3: Distribute and simplify:
76 - 4y + 3y = 64
-y = 64 - 76
-y = -12
Step 4: Multiply through by -1 to isolate y:
y = 12
Step 5: Substitute the value of y into Equation 2 to solve for x:
x + 12 = 19
x = 19 - 12
x = 7
Therefore, the two numbers are 7 and 12.