Introduction to Problem Solving: Videos & Practice Problems

Solving word problems involves understanding the problem, defining variables, and building equations like $P=2l+2w$ for perimeter. Using relationships such as length being four times width ($l=4w$), we substitute and simplify to find dimensions. Checking solutions ensures accuracy and reasonableness. This approach enhances skills in translating real-world scenarios into algebraic expressions, crucial for mastering polynomials, terms, and coefficients in various mathematical contexts.

Word Problem Solving Strategy

Solving word problems involving linear equations becomes manageable by following a systematic approach that enhances problem-solving confidence. Begin by thoroughly understanding the problem through careful reading, identifying key information, and defining variables clearly. For example, when dealing with a rectangular field where the length is four times the width and the perimeter is 500 yards, it helps to visualize the scenario by drawing a diagram and assigning variables such as length l and width w, with perimeter P.

Next, construct an equation that models the problem. The perimeter of a rectangle is calculated by adding twice the length and twice the width, expressed as the formula \(P = 2l + 2w\). Since the length is four times the width, substitute \(l = 4w\) into the perimeter equation, resulting in \(P = 2(4w) + 2w = 8w + 2w = 10w\). Given that the perimeter \(P\) is 500 yards, the equation becomes \$500 = 10w\(.

Solving for the width involves isolating the variable by dividing both sides by 10, yielding \)w = \frac{500}{10} = 50\(. Using the relationship between length and width, calculate the length as \)l = 4 \times 50 = 200\(. Thus, the field's dimensions are 50 yards in width and 200 yards in length.

It is crucial to verify the solution by substituting the values back into the original equation: \)500 = 2(200) + 2(50) = 400 + 100 = 500$, confirming the correctness of the solution. Additionally, assessing the reasonableness of the answer within the problem's context ensures that the dimensions make sense given the total perimeter.

This methodical process—understanding the problem, building an equation, solving it, stating the answer, and checking the solution—provides a reliable framework for tackling various word problems involving linear equations. Mastery of translating verbal statements into algebraic expressions and manipulating equations is essential for success in solving real-world problems efficiently.

Find the unknown numbers.
One number is nine less than another. Their sum is negative twenty-seven.

• A. $-18$ and $-27$
• B. $-18$ and $-9$
• C. $-27$ and $-9$
• D. $-9$ and $18$

Answer: (B)

Find the unknown number.
If half of a number is added to $\frac{2}{5}$, the result is the same as subtracting $\frac{1}{10}$ from the number.

• A. $-\frac{1}{5}$
• B. $-1$
• C. $1$
• D. $-\frac{3}{5}$

Answer: (C)

Patricia has $30$ meters of fencing to make a rectangular garden in her backyard. She wants the length to be $5$ meters more than the width. Complete steps of the word problem solving process to set up an equation Patricia could use to find the width of her rectangular fence.

• A. $30=2(w+5)+2w$
• B. $30=2(w+5)+w$
• C. $30=(w+5)+2w$
• D. $30=2(w-5)+2w$

Answer: (A)

Jordan is designing a picture frame for a poster. The perimeter of the frame is $80\mathrm{cm}$. The length is $12\mathrm{cm}$ longer than its width. Identify the dimensions of this poster.

• A. $L=46\mathrm{cm}$ and $w=34\mathrm{cm}$
• B. $L=23\mathrm{cm}$ and $w=17\mathrm{cm}$
• C. $L=52\mathrm{cm}$ and $w=28\mathrm{cm}$
• D. $L=26\mathrm{cm}$ and $w=14\mathrm{cm}$

Answer: (D)

Word Problem Solving Strategy Example 1

In a triangle, the sum of all interior angles always equals 180 degrees, which is a fundamental property used to solve for unknown angles. Consider a triangle with two known side lengths of 8 meters and 12 meters, and three angles related by specific expressions. One angle is denoted as x, another angle is 20 degrees more than x (expressed as x + 20°), and the angle between the two known sides is 30 degrees less than twice x, represented as 2x - 30°.

To find the values of these angles, set up an equation based on the angle sum property:

\[x + (x + 20) + (2x - 30) = 180\]

Combining like terms simplifies the equation to:

\[4x - 10 = 180\]

Adding 10 to both sides gives:

\[4x = 190\]

Dividing both sides by 4 yields the value of x:

\[x = 47.5^\circ\]

With x known, calculate the other angles by substituting back into their expressions. The angle 20 degrees more than x is:

\[x + 20 = 47.5^\circ + 20^\circ = 67.5^\circ\]

The angle 30 degrees less than twice x is:

\[2x - 30 = 2 \times 47.5^\circ - 30^\circ = 95^\circ - 30^\circ = 65^\circ\]

Verifying the solution, the sum of the angles is:

\[47.5^\circ + 67.5^\circ + 65^\circ = 180^\circ\]

This confirms the accuracy of the calculations. Understanding how to translate verbal descriptions of angle relationships into algebraic expressions and applying the triangle angle sum theorem is essential for solving such geometric problems efficiently.

Here’s what students ask on this topic:

What are the key steps to solve word problems in algebra?

To solve word problems in algebra effectively, follow these five key steps: First, understand the problem by reading it carefully and identifying what is being asked. Second, define and name the variables involved. Third, build an equation that models the problem using the relationships described. Fourth, solve the equation for the unknown variable(s). Fifth, state the answer clearly in the context of the problem and finally, check your solution to ensure it is correct and reasonable. This systematic approach helps break down complex problems into manageable parts, making it easier to translate real-world scenarios into algebraic expressions and solve them confidently.

How do you translate word problems into algebraic equations?

Translating word problems into algebraic equations involves identifying the variables and the relationships between them described in the problem. For example, if a problem states that the length of a rectangle is four times its width, you can express this as $l=4w$. Next, use known formulas, such as the perimeter of a rectangle $P=2l+2w$, and substitute the relationships to form an equation with one variable. This process of defining variables and expressing relationships algebraically is essential for solving word problems systematically.

How can substitution be used to solve equations with multiple variables in word problems?

Substitution is a powerful technique to solve equations with multiple variables by expressing one variable in terms of another and replacing it in the equation. For instance, if you know $l=4w$, you can substitute $4w$ for $l$ in the perimeter equation $P=2l+2w$. This reduces the equation to one variable, making it easier to solve. After finding the value of one variable, you can back-substitute to find the others. This method simplifies complex problems and is essential in solving word problems involving multiple unknowns.

Why is it important to check your solution in word problems?

Checking your solution in word problems is crucial to ensure both mathematical accuracy and contextual reasonableness. After solving the equation, substitute your values back into the original equation to verify correctness. Additionally, assess whether the solution makes sense in the real-world context. For example, if the perimeter of a field is 500 yards, a length of 600 yards would be unreasonable. This step helps catch errors in calculation or interpretation and confirms that your answer is valid and practical within the problem's scenario.

How do you build an equation for the perimeter of a rectangle in algebra?

The perimeter $P$ of a rectangle is the total distance around it, calculated by adding all sides. Since a rectangle has two lengths and two widths, the perimeter equation is $P=2l+2w$, where $l$ is the length and $w$ is the width. This formula is fundamental in word problems involving rectangular shapes, allowing you to relate the perimeter to the dimensions and solve for unknown variables.