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Consider the equation \( -\frac{1}{3} y+6=5 \) represented on the double number line. Use the double number line to decompose the variable expression and determine the solution to the equation. \( y= \)

Ask by Osborne Logan. in the United States
Jan 24,2025

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La solución de la ecuación \( -\frac{1}{3} y + 6 = 5 \) es \( y = 3 \).

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Beyond the Answer

To solve the equation \( -\frac{1}{3} y + 6 = 5 \), you can start by isolating \( y \). First, subtract 6 from both sides to get \( -\frac{1}{3} y = 5 - 6 \). This simplifies to \( -\frac{1}{3} y = -1 \). Now, to eliminate the negative sign and the fraction, multiply both sides by -3: \( y = (-1) \cdot (-3) \) which gives you \( y = 3 \). So, the solution to the equation is \( y = 3 \). The double number line can visually represent the relationship and help trace these changes step-by-step, making it easier to see how the equation balances out.

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PHIẾU: PHÉP CỘNG, TRỪ PHÂN SỐ HỌ VÀ TÊN:.............................LỚP 8A4 Bài 1. Thực hiện phép tính: a) \( \frac{x-5}{5}+\frac{1-x}{5} \) b) \( \frac{x-y}{8}+\frac{2 y}{8} \) c) \( \frac{x^{2}+1}{x-2}-\frac{1-2 x}{x-2} \) d) \( \frac{4 x+1}{3}-\frac{x-2}{3} \) e) \( \frac{4 x-1}{3 x^{2} y}-\frac{7 x-1}{3 x^{2} y} \) f) \( \frac{3 x+2 y}{x-y}-\frac{2 x+3 y}{x-y} \) d) \( \frac{5 x y^{2}-x^{2} y}{3 x y}-\frac{4 x y^{2}+x^{2} y}{3 x y} \) e) \( \frac{x+1}{a-b}+\frac{x-1}{a-b}-\frac{x+3}{a-b} \) f) \( \frac{5 x y-4 y}{2 x^{2} y^{3}}+\frac{3 x y+4 y}{2 x^{2} y^{3}} \) h) \( \frac{x^{2}+4}{x-2}+\frac{4 x}{2-x} \) i) \( \frac{2 x^{2}-x y}{x-y}+\frac{x y+y^{2}}{y-x}-\frac{2 y^{2}-x^{2}}{x-y} \) Bài 2: Thực hiện phép tính: a) \( \frac{2 x+4}{10}+\frac{2-x}{15} \) b) \( \frac{x^{2}}{x^{2}+3 x}+\frac{3}{x+3}+\frac{3}{x} \) c) \( \frac{2}{x+y}-\frac{1}{y-x}+\frac{-3 x}{x^{2}-y^{2}} \) d) \( \frac{4}{x+2}+\frac{2}{x-2}+\frac{5 x-6}{4-x^{2}} \); e) \( \frac{1-3 x}{2 x}+\frac{3 x-2}{2 x-1}+\frac{3 x-2}{2 x-4 x^{2}} \); f) \( \frac{x^{2}+2}{x^{3}-1}+\frac{2}{x^{2}+x+1}+\frac{1}{1-x} \) Bài 3. Làm tính trừ các phân thức sau: a) \( \frac{4 x+1}{3}-\frac{x-2}{3} \) b) \( \frac{4 x-1}{3 x^{2} y}-\frac{7 x-1}{3 x^{2} y} \) c) \( \frac{3 x+2 y}{x-y}-\frac{2 x+3 y}{x-y} \) Bài 4. Làm các phép tính a) \( \frac{x y-1}{2 x-y}-\frac{1-2 x^{2}}{y-2 x} \) b) \( \frac{3 x y^{2}+x^{2} y}{x^{2} y-x y^{2}}-\frac{3 x^{2} y+x y^{2}}{x y(x-y)} \) c) \( \frac{x+9}{x^{2}-9}-\frac{3}{x^{2}+3 x} \) Bài 5. Thực hiện phép tính a) \( \frac{5 x^{2}}{6 x-6 y}-\frac{2 x^{2}}{3 y-3 x} \) b) \( \frac{y}{x y-5 x^{2}}-\frac{25 x-15 y}{25 x^{2}-y^{2}} \) c) \( \frac{1}{2 x-3}-\frac{2}{2 x+3}-\frac{6}{4 x^{2}-9} \) Bài 6. Rút gọn rồi tính giá trị của biểu thức a) \( \frac{x+1}{x-1}-\frac{4 x}{x^{2}-1} \) với \( x=-2 \) b) \( \frac{1}{x y-x^{2}}-\frac{1}{y^{2}-x y} \) với \( x=-5 ; y=\frac{-1}{5} \) Bài 7. Chứng minh biểu thức sau không phụ thuộc vào x : \[ A=\frac{11 x}{2 x-3}-\frac{x-18}{3-2 x} \quad B=\frac{1}{x+1}-\frac{2 x}{x-1}+\frac{x+3}{x^{2}-1} \] Bài 8. Chứng minh biểu thức sau không phụ thuộc vào \( \mathrm{x}, \mathrm{y} \) : \[ D=\frac{2 x+1}{x-2 y}-\frac{5 y+2}{x-2 y}-\frac{y+1}{2 y-x} \quad K=\frac{x+y}{x}-\frac{x}{x-y}+\frac{y^{2}}{x^{2}-x y} \]
Algebra Vietnam Jan 24, 2025
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