The sum of five times a number and twice a second number is 18, but the difference of three times the first number and seven times the second number is -22. Find the two numbers using the elimination method.
5n+2m=18
3n-7m=-22
can you solve this?
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"The sum of five times a number and twice a second number is 18" ---->5x + 2y = 18
"the difference of three times the first number and seven times the second number is -22" ---> 3x - 7y = -22
1st times 7 -- 35x + 14y = 126
2nd times 2 -- 6x - 14y = -44
add them
41x = 82
x = 82/41= 2
into the 1st
5(2) + 2y = 18
10 + 2y = 18
2y = 8
y = 4
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To solve this problem using the elimination method, we need to set up a system of equations. Let's call the first number "x" and the second number "y".
From the problem statement, we can translate the given information into equations:
Equation 1: "The sum of five times a number and twice a second number is 18."
This can be written as: 5x + 2y = 18
Equation 2: "The difference of three times the first number and seven times the second number is -22."
This can be written as: 3x - 7y = -22
Now, we can solve this system of equations using the elimination method.
Step 1: Multiply both sides of Equation 1 by 3 and Equation 2 by 5 to eliminate the variables x.
15x + 6y = 54 (Equation 1 x 3)
15x - 35y = -110 (Equation 2 x 5)
Step 2: Subtract Equation 2 from Equation 1 to eliminate the variable x.
(15x + 6y) - (15x - 35y) = 54 - (-110)
15x + 6y - 15x + 35y = 54 + 110
41y = 164
Step 3: Solve for y.
Divide both sides of the equation by 41:
y = 164/41
y = 4
Step 4: Substitute the value of y=4 into Equation 1 (or Equation 2) and solve for x.
5x + 2(4) = 18
5x + 8 = 18
5x = 18 - 8
5x = 10
x = 10/5
x = 2
Therefore, the two numbers are x = 2 and y = 4.