Balancing Equations by Interchanging Numbers

The problem asks us to find which pair of numbers, when interchanged on the same side of the equation, makes the Left-Hand Side (LHS) equal to the Right-Hand Side (RHS).

The given equation is:

\(5 + 3 \times 6 - 4 \div 2 = 4 \times 3 - 10 \div 2 + 7\)

Evaluating the Original Equation

First, let's calculate the value of the original LHS and RHS using the order of operations (BODMAS/PEDMAS - Brackets, Orders/Exponents, Division and Multiplication, Addition and Subtraction). Division and Multiplication are done from left to right, as are Addition and Subtraction.

Calculate LHS:

\(LHS = 5 + 3 \times 6 - 4 \div 2\)

Perform Multiplication and Division:

\(3 \times 6 = 18\)

\(4 \div 2 = 2\)

Substitute back into LHS:

\(LHS = 5 + 18 - 2\)

Perform Addition and Subtraction from left to right:

\(5 + 18 = 23\)

\(23 - 2 = 21\)

So, the original LHS = 21.

Calculate RHS:

\(RHS = 4 \times 3 - 10 \div 2 + 7\)

Perform Multiplication and Division:

\(4 \times 3 = 12\)

\(10 \div 2 = 5\)

Substitute back into RHS:

\(RHS = 12 - 5 + 7\)

Perform Addition and Subtraction from left to right:

\(12 - 5 = 7\)

\(7 + 7 = 14\)

So, the original RHS = 14.

Originally, \(21 \neq 14\), so the equation is not balanced.

Testing the Options for Interchange

We need to test each option by interchanging the specified numbers on the same side (either LHS or RHS, wherever both numbers exist) and see if the modified side equals the original value of the other side (LHS = 14 or RHS = 21).

Option 1: Interchange 5 and 2

Interchange 5 and 2 on the LHS:

\(LHS_{new} = 2 + 3 \times 6 - 4 \div 5\)

\(LHS_{new} = 2 + 18 - 0.8\)

\(LHS_{new} = 20 - 0.8 = 19.2\)

\(19.2 \neq 14\). Let's try interchanging 5 and 2 on the RHS. Note that '5' is not present as a number on the RHS. The number 10 contains a 5, but the number itself is 10. Similarly, 2 is present as a number. If we interpret this as swapping the number 2 with the number 5 wherever they appear: swap 2 with 5 on RHS.

\(RHS_{new} = 4 \times 3 - 10 \div 5 + 7\)

\(RHS_{new} = 12 - 2 + 7\)

\(RHS_{new} = 10 + 7 = 17\)

\(17 \neq 21\). Option 1 does not balance the equation.

Option 2: Interchange 3 and 7

Interchange 3 and 7 on the LHS. Note that '7' is not present as a number on the LHS. Let's interchange 3 and 7 on the RHS.

\(RHS_{new} = 4 \times 7 - 10 \div 2 + 3\)

\(RHS_{new} = 28 - 5 + 3\)

\(RHS_{new} = 23 + 3 = 26\)

\(26 \neq 21\). Option 2 does not balance the equation.

Option 3: Interchange 6 and 4

Interchange 6 and 4 on the LHS:

\(LHS_{new} = 5 + 3 \times 4 - 6 \div 2\)

Perform Multiplication and Division:

\(3 \times 4 = 12\)

\(6 \div 2 = 3\)

Substitute back into LHS:

\(LHS_{new} = 5 + 12 - 3\)

Perform Addition and Subtraction:

\(LHS_{new} = 17 - 3 = 14\)

The new LHS is 14. This is equal to the original RHS (which was 14).

\(LHS_{new} = 14\)

\(RHS_{original} = 14\)

Since \(LHS_{new} = RHS_{original}\), interchanging 6 and 4 on the LHS makes the equation balanced.

Let's check if interchanging 6 and 4 on the RHS works. Note that '6' is not present as a number on the RHS. So, this interchange is not possible on the RHS.

Therefore, interchanging 6 and 4 on the LHS makes the equation balanced.

Option 4: Interchange 4 and 7

Interchange 4 and 7 on the LHS. Note that '7' is not present as a number on the LHS. Let's interchange 4 and 7 on the RHS.

\(RHS_{new} = 7 \times 3 - 10 \div 2 + 4\)

\(RHS_{new} = 21 - 5 + 4\)

\(RHS_{new} = 16 + 4 = 20\)

\(20 \neq 21\). Option 4 does not balance the equation.

Conclusion

Interchanging the numbers 6 and 4 on the LHS of the equation makes the LHS equal to the original RHS value, thereby balancing the equation.

The final answer is 6 and 4.

Revision Table: Original vs. Modified Equation

After interchanging 6 and 4 on the LHS, the equation becomes \(14 = 14\), which is true.

Additional Information: Order of Operations (BODMAS/PEDMAS)

To correctly evaluate mathematical expressions, we follow a specific order of operations. This order ensures that everyone gets the same result from the same expression. The common acronyms are BODMAS or PEDMAS.

• Brackets (or Parentheses)
• Orders (or Exponents/Powers/Roots)
• Division and Multiplication (done from left to right)
• Addition and Subtraction (done from left to right)

In the given equation, we performed multiplication and division before addition and subtraction, following this rule. When interchanging numbers, the operations and their positions remain the same, only the numbers change.