\frac{14 m-80}{16 m^2}-\frac{9 n^2+1}{4 m-3 n}
Answer (1)
Find a common denominator: 16 m 2 ( 4 m − 3 n ) .
Rewrite the fractions with the common denominator.
Combine the fractions into a single fraction.
Simplify the numerator by expanding and combining like terms: 8 m 2 ( 4 m − 3 n ) − 72 m 2 n 2 + 20 m 2 − 21 mn − 160 m + 120 .
The simplified expression is: 8 m 2 ( 4 m − 3 n ) − 72 m 2 n 2 + 20 m 2 − 21 mn − 160 m + 120 .
Explanation
Problem Analysis We are given the expression: 16 m 2 14 m − 80 − 4 m − 3 n 9 n 2 + 1 Our goal is to simplify this expression.
Finding Common Denominator First, let's find a common denominator for the two fractions. The common denominator is 16 m 2 ( 4 m − 3 n ) . We rewrite each fraction with this common denominator: 16 m 2 ( 4 m − 3 n ) ( 14 m − 80 ) ( 4 m − 3 n ) − 16 m 2 ( 4 m − 3 n ) ( 9 n 2 + 1 ) ( 16 m 2 )
Combining Fractions Now, we combine the fractions into a single fraction: 16 m 2 ( 4 m − 3 n ) ( 14 m − 80 ) ( 4 m − 3 n ) − ( 9 n 2 + 1 ) ( 16 m 2 )
Expanding the Numerator Next, we simplify the numerator by expanding the terms: 16 m 2 ( 4 m − 3 n ) ( 56 m 2 − 42 mn − 320 m + 240 ) − ( 144 m 2 n 2 + 16 m 2 ) 16 m 2 ( 4 m − 3 n ) 56 m 2 − 42 mn − 320 m + 240 − 144 m 2 n 2 − 16 m 2 16 m 2 ( 4 m − 3 n ) 40 m 2 − 42 mn − 320 m + 240 − 144 m 2 n 2
Factoring the Numerator We can factor out a common factor of 2 from the numerator: 16 m 2 ( 4 m − 3 n ) 2 ( 20 m 2 − 21 mn − 160 m + 120 − 72 m 2 n 2 ) 8 m 2 ( 4 m − 3 n ) 20 m 2 − 21 mn − 160 m + 120 − 72 m 2 n 2
Final Simplified Expression The simplified expression is: 8 m 2 ( 4 m − 3 n ) 20 m 2 − 21 mn − 160 m + 120 − 72 m 2 n 2 We can also write it as: 8 m 2 ( 4 m − 3 n ) − 72 m 2 n 2 + 20 m 2 − 21 mn − 160 m + 120
Alternative Form Alternatively, we can express the simplified expression as: 8 m 2 ( 4 m − 3 n ) − 8 m 2 ( 9 n 2 + 1 ) + ( 4 m − 3 n ) ( 7 m − 40 )
Examples
Simplifying rational expressions is a fundamental skill in algebra, with applications in various fields. For instance, in physics, when dealing with complex circuits, simplifying expressions involving resistance and impedance can make calculations more manageable. Similarly, in economics, simplifying cost functions can help in optimizing production processes. By mastering these techniques, students gain a valuable tool for solving real-world problems efficiently.
