yea it sais, three times one number added to another number is 18. twice the first number minus the other number is twelve.
find the numbers
x=first number
y=second number
3*x + y = 18
2*x - y = 12
You need to eliminate one of the variables and solve for the remaining one. In this problem, you can see that if the two equations are added together, the y variable is eliminated:
adding:
3x + y = 18
2x - y = 12
---------------
5x + 0y = 30
5x = 30
x=6
plug the known x value into either equation to solve for y.
taking the first one...
3*6 + y = 18
rearrange to get
y = 0
hey thx man, i messed up that one cause i though added to another number ment it goes before it
To solve this problem, we can set up a system of equations based on the given information.
Let's denote the two numbers as "x" and "y".
From the first statement, "Three times one number added to another number is 18," we can write the equation:
3x + y = 18.
From the second statement, "Twice the first number minus the other number is twelve," we can write the equation:
2x - y = 12.
Now we have a system of two equations with two unknowns. We can solve the system using elimination or substitution method.
Let's use the elimination method to solve the system:
Multiply the second equation by 3 to eliminate the variable "y":
3(2x - y) = 3(12)
6x - 3y = 36.
Now we have the modified system of equations:
3x + y = 18,
6x - 3y = 36.
Next, we can add the two equations together to eliminate the variable "y":
(3x + y) + (6x - 3y) = 18 + 36
9x - 2y = 54.
Now we have the new equation:
9x - 2y = 54.
From here, we can solve for "x" or "y". Let's solve for "x":
9x - 2y = 54
9x = 2y + 54
x = (2y + 54) / 9
x = (2/9)y + 6.
Now we need to find the values of "x" and "y" that satisfy both equations.
One way to proceed is to substitute the expression for "x" in terms of "y" in one of the original equations. Let's use the first equation:
3x + y = 18.
Substitute (2/9)y + 6 for "x" in the equation:
3((2/9)y + 6) + y = 18
(6/9)y + 18 + y = 18
(6/9 + 1)y = 18 - 18
(6/9 + 9/9)y = 0
(15/9)y = 0.
To get rid of the fraction, we can multiply both sides of the equation by (9/15):
(9/15)(15/9)y = (9/15)(0)
y = 0.
Now that we have the value of "y", we can substitute it back into one of the original equations to solve for "x". Let's use the second equation:
2x - y = 12.
Substitute 0 for "y" in the equation:
2x - 0 = 12
2x = 12
x = 12/2
x = 6.
Therefore, the two numbers are x = 6 and y = 0.