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 $l=4w$, substitute to solve for variables. Check solutions by verifying equations and context. This approach enhances skills in translating real-world scenarios into algebraic expressions, crucial for mastering polynomials, terms, coefficients, and exponents in standard form and scientific notation.

Word Problem Solving Strategy

Solving word problems involving linear equations becomes manageable by following a systematic approach. Begin by thoroughly understanding the problem, which involves carefully reading the question, visualizing the scenario—often by drawing a diagram—and identifying the variables involved. For example, consider a rectangular field where the length is four times the width, and the perimeter is 500 yards. Assign variables such as l for length, w for width, and p for perimeter to clearly represent the quantities.

Next, construct an equation that models the problem. The perimeter p of a rectangle is calculated by adding twice the length and twice the width, expressed as \(p = 2l + 2w\). Since the length is four times the width, this relationship can be written as \(l = 4w\). Substituting this into the perimeter equation gives \(p = 2(4w) + 2w\), which simplifies to \(p = 8w + 2w = 10w\). Knowing the perimeter is 500 yards, substitute \(p = 500\) to get \$500 = 10w\(.

Solving for the width involves isolating w by dividing both sides by 10, resulting in \)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.

Finally, verify the solution by substituting the values back into the original equation: \)500 = 2(200) + 2(50)\(, which simplifies to \)500 = 400 + 100 = 500$. This confirms the solution satisfies the equation. Additionally, assess the reasonableness of the answer in context; since both dimensions are less than the total perimeter, the values are logical. This methodical process—understanding the problem, modeling with equations, solving, stating the answer, and checking—builds confidence and accuracy in solving word problems involving linear equations.

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. $-\frac15$
• B. $-1$
• C. $1$
• D. $-\frac35$

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 $1$ & $2$ of the word problem solving process to set up an equation Patricia could use to find the width of her rectangular fence.

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

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

Answer: (D)

Word Problem Solving Strategy Example 1

In a triangle, the sum of all interior angles is always 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, which can be written 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:

\[x = \frac{190}{4} = 47.5^\circ\]

With x known, calculate the other angles:

• The angle 20 degrees more than x is:

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

• The angle 30 degrees less than twice x is:

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

Verifying the solution by summing all angles confirms the accuracy:

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

This problem highlights the importance of understanding angle relationships and the triangle angle sum theorem. By translating verbal descriptions into algebraic expressions and solving linear equations, one can determine unknown angles in geometric figures effectively. This approach is essential in geometry, trigonometry, and real-world applications involving triangular shapes.

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, drawing diagrams if needed, and identifying variables. Second, build an equation that models the problem using the relationships described. Third, solve the equation for the unknown variable(s). Fourth, state the answer clearly in the context of the problem. Finally, check your solution by substituting values back into the equation and ensuring the answer makes sense in the real-world context. This systematic approach helps break down complex problems into manageable parts, improving accuracy and confidence in problem-solving.

How do you translate word problems into algebraic equations?

Translating word problems into algebraic equations involves identifying the variables and relationships described in the problem. For example, if a problem states that the length of a rectangle is four times its width, you can represent 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 requires careful reading and practice to accurately convert words into mathematical expressions.

How can I check if my solution to a word problem is correct?

After solving a word problem, check your solution in two ways. First, substitute your values back into the original equation to verify the equality holds true. For example, if you found $w=50$ and $l=200$, plug these into the perimeter formula $P=2l+2w$ to confirm $500=2\times 200+2\times 50$ is true. Second, assess if the solution is reasonable in context; for instance, dimensions should be positive and consistent with the problem's conditions. This dual check ensures accuracy and practical validity.

Why is it important to define variables clearly in word problems?

Defining variables clearly is crucial because it helps organize information and simplifies the process of building equations. Assigning meaningful variable names, like $l$ for length and $w$ for width, makes it easier to track relationships and substitutions. Clear variables reduce confusion, especially when multiple quantities are involved, and facilitate communication of your solution. This clarity is essential for solving complex problems accurately and efficiently.

How do I handle multiple variables in word problems?

When a word problem involves multiple variables, start by expressing some variables in terms of others using the relationships given. For example, if length $l$ is four times width $w$, write $l=4w$. Substitute this into the main equation to reduce the number of variables. This method allows you to solve for one variable at a time. Sometimes, you may need to create and solve a system of equations if multiple relationships exist. This approach simplifies complex problems into solvable equations.