Assigntment if Soluing Reallife Problems with simultanecus Equations (1) A schcol buys 5 book and 3 pens for \( k 45 \). Ansther schoul byys 3 books and 2 pens for \( k 29.00 \) Find the price of each bcok and pen. (2) A father is iwice as old as his son. In 5 years the sum of the ages will be 65 . Find their currer ages. (3) Find twu numbers whase sumand differenc is 4 .

**(1)** Let \( b \) be the price of one book and \( p \) the price of one pen. We are given: \[ \begin{aligned} 5b + 3p &= 45, \\ 3b + 2p &= 29. \end{aligned} \] *Step 1. Multiply the equations to eliminate one variable.* The least common multiple of 3 and 2 is 6. Multiply the first equation by 2 and the second equation by 3: \[ \begin{aligned} 2(5b + 3p) &= 2(45) \quad \Rightarrow \quad 10b + 6p = 90, \\ 3(3b + 2p) &= 3(29) \quad \Rightarrow \quad 9b + 6p = 87. \end{aligned} \] *Step 2. Subtract the second equation from the first to solve for \( b \).* \[ (10b + 6p) - (9b + 6p) = 90 - 87 \quad \Rightarrow \quad b = 3. \] *Step 3. Substitute \( b = 3 \) into one of the original equations to solve for \( p \). Using \( 3b + 2p = 29 \):* \[ 3(3) + 2p = 29 \quad \Rightarrow \quad 9 + 2p = 29 \quad \Rightarrow \quad 2p = 20 \quad \Rightarrow \quad p = 10. \] **Answer for (1):** The price of one book is \( k3 \) and the price of one pen is \( k10 \). --- **(2)** Let \( s \) be the current age of the son. Then the father's current age is \( 2s \). *Step 1. Express the ages in 5 years:* \[ \text{Son's age in 5 years} = s + 5, \quad \text{Father's age in 5 years} = 2s + 5. \] *Step 2. The sum of their ages in 5 years is given as 65:* \[ (s + 5) + (2s + 5) = 65. \] *Step 3. Simplify and solve for \( s \):* \[ 3s + 10 = 65 \quad \Rightarrow \quad 3s = 55 \quad \Rightarrow \quad s = \frac{55}{3}. \] *Step 4. Find the father's current age:* \[ 2s = 2 \times \frac{55}{3} = \frac{110}{3}. \] **Answer for (2):** The current age of the son is \( \frac{55}{3} \) years (approximately \( 18.\overline{3} \) years) and the current age of the father is \( \frac{110}{3} \) years (approximately \( 36.\overline{6} \) years). --- **(3)** Let the two numbers be \( x \) and \( y \). We are given: \[ \begin{aligned} x + y &= 4, \\ x - y &= 4. \end{aligned} \] *Step 1. Add the two equations to solve for \( x \):* \[ (x+y) + (x-y) = 4 + 4 \quad \Rightarrow \quad 2x = 8 \quad \Rightarrow \quad x = 4. \] *Step 2. Substitute \( x = 4 \) into \( x + y = 4 \) to solve for \( y \):* \[ 4 + y = 4 \quad \Rightarrow \quad y = 0. \] **Answer for (3):** The two numbers are \( 4 \) and \( 0 \).