2/x + 3/y = 7. 1/x-4/y=-1. How to determine the values of. X and y
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We have a system of equations with fractions. The first equation is 2 over x plus 3 over y equals 7. The second equation is 1 over x minus 4 over y equals negative 1. To solve this system, we'll use substitution. Let u equal 1 over x and v equal 1 over y. This transforms our system into linear equations: 2u plus 3v equals 7, and u minus 4v equals negative 1.
Now we'll solve this linear system using the elimination method. We have equation 1: 2u plus 3v equals 7, and equation 2: u minus 4v equals negative 1. To eliminate u, we multiply equation 2 by 2, giving us 2u minus 8v equals negative 2. Now we subtract this new equation from equation 1. This gives us 11v equals 9, so v equals 9 over 11.
Now we substitute v equals 9 over 11 back into equation 2. We get u minus 4 times 9 over 11 equals negative 1. This simplifies to u minus 36 over 11 equals negative 1. Solving for u, we get u equals negative 1 plus 36 over 11, which equals 25 over 11. Now we convert back to x and y. Since u equals 1 over x, we have x equals 11 over 25. Since v equals 1 over y, we have y equals 11 over 9.
Let's verify our solution. We found x equals 11 over 25 and y equals 11 over 9. For equation 1, we substitute: 2 over x plus 3 over y becomes 2 times 25 over 11 plus 3 times 9 over 11, which equals 50 over 11 plus 27 over 11, giving us 77 over 11, which equals 7. This checks out. For equation 2, we get 25 over 11 minus 36 over 11, which equals negative 11 over 11, or negative 1. Both equations are satisfied, confirming our solution is correct.
We have successfully solved the system of equations. Our final answer is x equals 11 over 25 and y equals 11 over 9. To summarize our method: we used substitution to transform the fractional equations into a linear system, applied the elimination method to solve for the new variables, then converted back to find x and y. We verified our solution by substituting back into the original equations. In decimal form, x is approximately 0.44 and y is approximately 1.22. This substitution technique is very useful for solving systems with fractions or other complex expressions.