find twonumbers
such that the sum of twice the first and thric the second is 92 and four times the exceeds seven times the second by 2
2 x + 3 y = 92
4 x - 7 y = 2
4 x + 6 y = 184
4 x - 7 y = 2
---------------------
13 y = 182
y = 14 etc
Damon why did you pass me up? Don't you like me amymore? :0(
I do not have a politically correct answer to your preamble question.
To find two numbers that satisfy the given conditions:
Let's assume the first number as "x" and the second number as "y."
According to the first condition:
"Twice the first number (2x) plus three times the second number (3y) is equal to 92."
This can be written as the equation: 2x + 3y = 92.
According to the second condition:
"Four times the first number (4x) exceeds seven times the second number (7y) by 2."
This can be written as the equation: 4x = 7y + 2.
Now, we have a system of two equations:
1) 2x + 3y = 92
2) 4x = 7y + 2
We can solve this system of equations using different methods. One way is to use the substitution method.
From equation 2, we can isolate x:
4x = 7y + 2
Divide both sides by 4:
x = (7y + 2) / 4
Now substitute this value of x in equation 1:
2((7y + 2) / 4) + 3y = 92
Simplify and solve for y:
(7y + 2) / 2 + 3y = 92
Multiply through by 2 to eliminate the fraction:
7y + 2 + 6y = 184
Combine like terms:
13y + 2 = 184
Subtract 2 from both sides:
13y = 182
Divide both sides by 13:
y = 14
Now substitute the value of y back into any of the original equations. Let's use equation 2:
4x = 7(14) + 2
4x = 98 + 2
4x = 100
Divide both sides by 4:
x = 25
Therefore, the two numbers that satisfy the given conditions are x = 25 and y = 14.