Understanding the Number Word Problem

The problem provides us with two statements relating two unknown numbers. We need to use these statements to set up mathematical equations and then solve for the value of the first number.

Let's define our variables:

• Let the first number be represented by x.
• Let the second number be represented by y.

Setting Up the Equations from the Problem Statements

We translate each statement into an algebraic equation:

1. "One-fourth of a number is equal to three-eighth of another number."
   This translates to: $\frac{1}{4}x = \frac{3}{8}y$
2. "If 30 is added to the first number, then it becomes six times that of the second number."
   This translates to: $x + 30 = 6y$

Now we have a system of two linear equations with two variables:

Equation 1: $\frac{x}{4} = \frac{3y}{8}$

Equation 2: $x + 30 = 6y$

Solving the System of Equations

We can use the substitution method or the elimination method to solve this system. Let's use substitution.

From Equation 1, we can express x in terms of y:

$\frac{x}{4} = \frac{3y}{8}$

Multiply both sides by 4:

$x = 4 \times \frac{3y}{8}$

$x = \frac{12y}{8}$

Simplify the fraction:

$x = \frac{3y}{2}$

Now, substitute this expression for x into Equation 2:

$x + 30 = 6y$

$\frac{3y}{2} + 30 = 6y$}

To eliminate the fraction, multiply the entire equation by 2:

$2 \times \left(\frac{3y}{2} + 30\right) = 2 \times (6y)$

$3y + 60 = 12y$}

Now, we need to solve for y. Subtract $3y$ from both sides:

$60 = 12y - 3y$

$60 = 9y$}

Divide both sides by 9:

$y = \frac{60}{9}$}

Simplify the fraction:

$y = \frac{20}{3}$

Finding the First Number (x)

We found that $y = \frac{20}{3}$. Now we can substitute this value back into the expression for x that we derived from Equation 1 ($x = \frac{3y}{2}$):

$x = \frac{3}{2} \times y$}

$x = \frac{3}{2} \times \frac{20}{3}$}

Multiply the numerators and the denominators:

$x = \frac{3 \times 20}{2 \times 3}$}

$x = \frac{60}{6}$}

Divide to find the value of x:

$x = 10$

So, the first number is 10.

Verifying the Answer

Let's check if our numbers ($x=10$ and $y=\frac{20}{3}$) satisfy the original conditions:

Condition 1: One-fourth of the first number is equal to three-eighth of the second number.

Left side: $\frac{1}{4}x = \frac{1}{4} \times 10 = \frac{10}{4} = \frac{5}{2}$

Right side: $\frac{3}{8}y = \frac{3}{8} \times \frac{20}{3} = \frac{3 \times 20}{8 \times 3} = \frac{60}{24} = \frac{5}{2}$

The first condition is satisfied: $\frac{5}{2} = \frac{5}{2}$.

Condition 2: If 30 is added to the first number, it becomes six times the second number.

Left side: $x + 30 = 10 + 30 = 40$

Right side: $6y = 6 \times \frac{20}{3} = \frac{120}{3} = 40$

The second condition is satisfied: $40 = 40$.

Both conditions are met, confirming that our calculated value for the first number, 10, is correct.

Revision Table: Key Steps

Additional Information: Solving Systems of Equations

A system of linear equations is a set of two or more linear equations involving the same variables. The solution to a system is the set of values for the variables that satisfy all equations simultaneously.

Common methods for solving systems of two linear equations include:

• Substitution Method: Solve one equation for one variable, then substitute that expression into the other equation. This reduces the system to a single equation with one variable.
• Elimination Method: Multiply one or both equations by constants so that the coefficients of one variable are opposites. Then, add the equations together to eliminate that variable, resulting in a single equation with one variable.
• Graphical Method: Graph both equations on the same coordinate plane. The point of intersection of the lines represents the solution to the system. This method is useful for visualization but may not provide exact solutions if the intersection point has non-integer coordinates.

Word problems often require setting up a system of equations based on the relationships described in the text, just like we did in this problem.