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HAHD-001 普通用户投稿 1</s>In the expression ( frac{x}{2} + frac{y}{3} = frac{z}{5} ), where ( x, y, z ) are positive integers, find the value of ( x + y + z ).To solve the equation ( frac{x}{2} + frac{y}{3} = frac{z}{5} ), we can first find a common denominator to rewrite the equation with all terms on the left-hand side: [ frac{2x + 3y}{6} = frac{5z}{15} ] Since ( 6 ) and ( 15 ) are both divisible by 3, we can divide both sides of the equation by 3 to simplify: [ frac{2x + 3y}{2} = frac{5z}{5} ] This simplifies to: [ 2x + 3y = 5z ] Now, we have a linear equation with integer coefficients. To find integer solutions for ( x, y, z ), we can use the Diophantine equation approach. Since the sum of the coefficients on the right-hand side (the terms with ( x ) and ( y )) is ( 2 + 3 = 5 ), we can try to find a value for ( z ) such that when multiplied by 5, the result has a 0 in the tens place (to ensure that ( 2x + 3y ) can be 5 times an integer). The easiest choice for ( z ) is ( z = 10 ), because when multiplied by 5, it results in ( 50 ), which is convenient for balancing the equation. So, we substitute ( z = 10 ) into the equation: [ 2x + 3y = 5 cdot 10 ] [ 2x + 3y = 50 ] Now, we can use trial and error to find integer values for ( x ) and ( y ) that satisfy the equation. One such solution is ( x =

2010年5月7日119 分钟