Kya aap section 1 se 15 tak ke tamam linear equations aur word problems ke step-by-step solutions provide kar sakte hain?

Question

Ahsan Malik · Accepted Answer

Yeh topics **PRC 2 (Quantitative Analysis for Business)** ke hain, jo mere domain (PRC 1 Fundamentals of Accounting) se baahar hain. Aapki aasani ke liye is pure page ke tamam sawalat ka direct step-by-step solution yeh raha:

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### **Section 1: Solve Linear Equations (1 to 8)**

**1.** $3x - 12 + 4x = 6x + 6$  
$7x - 12 = 6x + 6 \implies 7x - 6x = 6 + 12 \implies \mathbf{x = 18}$

**2.** $3(2y - 4) + 3y = 5(y - 4)$  
$6y - 12 + 3y = 5y - 20 \implies 9y - 12 = 5y - 20$  
$9y - 5y = -20 + 12 \implies 4y = -8 \implies \mathbf{y = -2}$

**3.** $3x - 2(2x - 5) = 2(x + 3) - 8$  
$3x - 4x + 10 = 2x + 6 - 8 \implies -x + 10 = 2x - 2$  
$-x - 2x = -2 - 10 \implies -3x = -12 \implies \mathbf{x = 4}$

**4.** $3x + \frac{1}{5} = 2 - x$  
$4x = 2 - \frac{1}{5} = \frac{9}{5} \implies x = \frac{9}{20} \quad (	ext{ya } \mathbf{0.45})$

**5.** $\frac{x}{4} + \frac{x}{6} = \frac{x}{2} + \frac{3}{4}$  
Puri equation ko $12$ (LCM) se multiply karein:  
$3x + 2x = 6x + 9 \implies 5x = 6x + 9 \implies \mathbf{x = -9}$

**6.** $\frac{y + 4}{2y - 3} = \frac{2}{5}$  
Cross multiply karein:  
$5(y + 4) = 2(2y - 3) \implies 5y + 20 = 4y - 6 \implies \mathbf{y = -26}$

**7.** $x - \left(2x - \frac{3x - 4}{7}ight) = \frac{4x - 27}{3} - 3$  
LHS solve karein: $-x + \frac{3x - 4}{7} = \frac{-7x + 3x - 4}{7} = \frac{-4x - 4}{7}$  
RHS solve karein: $\frac{4x - 27 - 9}{3} = \frac{4x - 36}{3}$  
Cross multiply: $3(-4x - 4) = 7(4x - 36) \implies -12x - 12 = 28x - 252$  
$40x = 240 \implies \mathbf{x = 6}$

**8.** $0.3x + 0.4 = 0.28x + 1.16$  
$0.30x - 0.28x = 1.16 - 0.40 \implies 0.02x = 0.76 \implies x = \frac{0.76}{0.02} \implies \mathbf{x = 38}$

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### **Question 9: Solution Points (Ordered Pairs)**

Equation mein pairs $(x, y)$ put karke check karein ke dono sides barabar aati hain ya nahi:

*   **(a)** $3x - y = -8$  
    *   $(-2, 2) ightarrow 3(-2) - 2 = -8$ **(Yes)**  
    *   $(-1, 5) ightarrow 3(-1) - 5 = -8$ **(Yes)**  
    *   *Solutions:* **$(-2, 2)$** aur **$(-1, 5)$**
*   **(b)** $2x - y = -1$  
    *   $(0, 1) ightarrow 2(0) - 1 = -1$ **(Yes)**  
    *   $(2, 5) ightarrow 2(2) - 5 = -1$ **(Yes)**  
    *   $(-2, -3) ightarrow 2(-2) - (-3) = -1$ **(Yes)**  
    *   *Solutions:* **$(0, 1)$**, **$(2, 5)$**, aur **$(-2, -3)$**
*   **(c)** $2x + y = 4$  
    *   $(0, 4) ightarrow 2(0) + 4 = 4$ **(Yes)**  
    *   *Solution:* **$(0, 4)$**

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### **Question 10: Three Ordered Pairs**

*   **(a)** $y - x = 3 \implies y = x + 3$  
    $x$ ki values $0, 1, 2$ rakhne par pairs: **$(0, 3)$, $(1, 4)$, $(2, 5)$**
*   **(b)** $3x + 2y = -6 \implies 2y = -6 - 3x$  
    *   $x = 0 \implies y = -3 ightarrow (0, -3)$  
    *   $x = -2 \implies y = 0 ightarrow (-2, 0)$  
    *   $x = 2 \implies y = -6 ightarrow (2, -6)$  
    *   Pairs: **$(0, -3)$, $(-2, 0)$, $(2, -6)$**

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### **Word Problems (11 to 15)**

*   **11.** Chaar consecutive odd numbers: $x, x+2, x+4, x+6$  
    $4x + 12 = 56 \implies 4x = 44 \implies x = 11$  
    Numbers hain $11, 13, 15, 17$. Greatest number: **17**

*   **12.** Let the number be $x$:  
    $2x + 8 = 50 \implies 2x = 42 \implies \mathbf{x = 21}$

*   **13.** Let Saima's age be $S$:  
    $5 = \frac{1}{4}S + 2 \implies \frac{1}{4}S = 3 \implies S = 3 	imes 4 = \mathbf{12	ext{ years}}$

*   **14.** Do numbers: $x$ aur $7x$  
    $(x + 7x) + 14 = 38 \implies 8x + 14 = 38 \implies 8x = 24 \implies x = 3$  
    Numbers hain: **3** aur **21** ($7 	imes 3$)

*   **15.** Ahsan = $A$, Fahad = $F$:  
    $A = 3F$  
    3 saal baad: $A + 3 = 2(F + 3) \implies 3F + 3 = 2F + 6 \implies F = 3$  
    $A = 3(3) = 9$  
    Ahsan ki age: **9 years**, Fahad ki age: **3 years**

[[SUGGEST_GROUP:MATHEMATICS]]