Solving the Equation: 5/x+y-2/x-y=-1 and 15/x+y+7/x-y=10 using Cross Multiplication
In this article, we will solve a system of linear equations using the cross multiplication method. The equations are:
Equation 1: 5/x + y - 2/x - y = -1 Equation 2: 15/x + y + 7/x - y = 10
Step 1: Simplify the Equations
Let's simplify the equations by combining like terms:
Equation 1: (5 - 2)/x + y(1 - 1) = -1 => 3/x = -1 Equation 2: (15 + 7)/x + y(1 - 1) = 10 => 22/x = 10
Step 2: Write the Equations in Standard Form
Now, let's write the equations in standard form, where the variables are on one side and the constants are on the other side:
Equation 1: 3/x = -1 Equation 2: 22/x = 10
Step 3: Apply Cross Multiplication
We will now apply cross multiplication to solve for x:
Equation 1: 3/x = -1 => 3 = -x Equation 2: 22/x = 10 => 22 = 10x
Now, let's multiply the two equations:
3 × 22 = -x × 10x 66 = -10x^2
Step 4: Solve for x
Now, let's solve for x:
-10x^2 = 66 x^2 = -66/10 x^2 = -6.6 x = ±√(-6.6) (not a real solution)
Since we couldn't find a real solution for x, let's go back to the original equations and try to solve for y.
Step 5: Solve for y
We will use Equation 1 to solve for y:
5/x + y - 2/x - y = -1 5/x - 2/x = -1 (5 - 2)/x = -1 3/x = -1 x = -3
Now, let's substitute x = -3 into Equation 1:
5/-3 + y - 2/-3 - y = -1 -5/3 + y + 2/3 - y = -1 -5/3 + 2/3 = -1 -3/3 = -1 -1 = -1 (True)
Now that we have found x, let's find y. We can use either Equation 1 or Equation 2 to find y. Let's use Equation 2:
15/-3 + y + 7/-3 - y = 10 -15/3 + y - 7/3 - y = 10 -15/3 - 7/3 = 10 -22/3 = 10 y = 10 + 22/3 y = (30 + 22)/3 y = 52/3
Therefore, the solution to the system of equations is x = -3 and y = 52/3.
