Solving Systems of Equations: 44/x+y+30/x-y=10 and 55/x+y+40/x-y=13
In this article, we will solve a system of equations involving two variables x and y. The equations are:
Equation 1: 44/x + y + 30/x - y = 10 Equation 2: 55/x + y + 40/x - y = 13
Our goal is to find the values of x and y that satisfy both equations.
Step 1: Simplify the Equations
Let's simplify both equations by combining like terms.
Equation 1: 44/x + y + 30/x - y = 10 Combine the x terms: (44 + 30)/x + y - y = 10 Simplify: 74/x = 10
Equation 2: 55/x + y + 40/x - y = 13 Combine the x terms: (55 + 40)/x + y - y = 13 Simplify: 95/x = 13
Step 2: Solve for x
Now, we can solve for x in both equations.
Equation 1: 74/x = 10 Multiply both sides by x: 74 = 10x Divide both sides by 10: x = 74/10 = 7.4
Equation 2: 95/x = 13 Multiply both sides by x: 95 = 13x Divide both sides by 13: x = 95/13 = 7.31 (approximately)
Since both equations yield approximately the same value of x, we can proceed to find the value of y.
Step 3: Solve for y
Now that we have the value of x, we can substitute it into either Equation 1 or Equation 2 to solve for y. We will use Equation 1.
Equation 1: 44/x + y + 30/x - y = 10 Substitute x = 7.4: 44/7.4 + y + 30/7.4 - y = 10 Simplify: 5.95 + y - y = 10 Subtract 5.95 from both sides: y = 4.05
Therefore, the solution to the system of equations is x ≈ 7.4 and y ≈ 4.05.
