Odpowiedź:
[tex]\dfrac{15x+6y}{3}-\dfrac{5x+10y}{5}=\dfrac{\not3^1(5x+2y)}{\not3_{1}}-\dfrac{\not5^1(x+2y)}{\not5_{1}}=5x+2y-(x+2y)=\\\\=5x+2y-x-2y=4x\\\\\\lub\ \ II\ \ spos\'ob\\\\\frac{15x+6y}{3}-\frac{5x+10y}{5}=\frac{15x}{3}+\frac{6y}{3}-\frac{5x}{5}-\frac{10y}{5}=5x+2y-x-2y=4x[/tex]
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5x+6y-x+10y= 4x+16y
Odpowiedź:
[tex]\dfrac{15x+6y}{3}-\dfrac{5x+10y}{5}=\dfrac{\not3^1(5x+2y)}{\not3_{1}}-\dfrac{\not5^1(x+2y)}{\not5_{1}}=5x+2y-(x+2y)=\\\\=5x+2y-x-2y=4x\\\\\\lub\ \ II\ \ spos\'ob\\\\\frac{15x+6y}{3}-\frac{5x+10y}{5}=\frac{15x}{3}+\frac{6y}{3}-\frac{5x}{5}-\frac{10y}{5}=5x+2y-x-2y=4x[/tex]