Two numbers. The first plus twice the second equals 11. Twice the first plus three times the second equals 50. What are the numbers?
x + 2y = 11
2x + 3y = 50
double the first , then subtract to get y
Once you have that , plug into the first to get x
double the first EQUATION , then subtract THE 2ND EQUATION to get y
2x + 4y - 2x - 3y = 22 - 50
y = -28
substitute into the 1st equation to find x
To solve this problem, we can create a system of linear equations based on the given information. Let's use the variables "x" and "y" to represent the first and second numbers, respectively.
According to the problem, the first plus twice the second equals 11. So, we can write the first equation as:
x + 2y = 11 -- Equation 1
The second piece of information states that twice the first plus three times the second equals 50. Therefore, the second equation is:
2x + 3y = 50 -- Equation 2
To solve this system of equations, we can use the method of elimination or substitution. I'll use the method of elimination in this example.
Step 1: Multiply both sides of Equation 1 by 2 to eliminate the coefficient of "x" in Equation 2.
2(x + 2y) = 2(11)
2x + 4y = 22 -- Equation 3
Step 2: Now, we can eliminate the variable "x" by subtracting Equation 3 from Equation 2.
(2x + 3y) - (2x + 4y) = 50 - 22
The x term cancels out, and the equation simplifies to:
0x - y = 28
-y = 28
Step 3: Solve for "y" by multiplying both sides by -1.
y = -28
Step 4: Substitute the value of "y" (-28) into any of the original equations. I'll use Equation 1:
x + 2(-28) = 11
x - 56 = 11
Step 5: Solve for "x" by adding 56 to both sides.
x = 11 + 56
x = 67
Therefore, the first number (x) is 67, and the second number (y) is -28.