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 ($l=4w$), substitution simplifies multivariable polynomials into solvable linear equations. Checking solutions ensures numerical coefficients and terms are reasonable within the problem's context. This systematic approach enhances problem-solving skills by translating real-world scenarios into algebraic expressions, reinforcing concepts like variables, coefficients, and equation solving in standard form.

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 reading it carefully, 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 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 = 50\( yards.

With the width determined, find the length by substituting back into the relationship \)l = 4w\(, yielding \)l = 4 \times 50 = 200\( yards. Thus, the field's dimensions are 200 yards in length and 50 yards in width.

Finally, verify the solution by substituting the values back into the original perimeter equation: \)500 = 2(200) + 2(50)\(, which simplifies to \)500 = 400 + 100 = 500$. This confirms the solution is mathematically correct. Additionally, assess the reasonableness of the answer in context; since both dimensions are less than the total perimeter, the values are logical and consistent with the problem's conditions.

By breaking down word problems into understanding, modeling, solving, stating, and checking, you can confidently tackle a variety of real-world scenarios involving linear equations. This methodical process not only simplifies complex problems but also ensures accuracy and clarity in your solutions.

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 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, 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:

\[x = 47.5^\circ\]

With x known, calculate the other angles by substituting back into their expressions:

Angle 2: \[x + 20 = 47.5^\circ + 20^\circ = 67.5^\circ\]

Angle 3 (between the two known sides): \[2x - 30 = 2 \times 47.5^\circ - 30^\circ = 65^\circ\]

Verifying the solution by summing all angles confirms the accuracy:

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

This problem illustrates how algebraic expressions can represent geometric relationships within triangles, and how applying the triangle angle sum theorem allows for solving unknown angles efficiently. Understanding these connections is essential for solving more complex geometric problems involving side lengths and angle measures.

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. Finally, check your solution by substituting values back into the original equation and ensuring the answer makes sense in the real-world context. This systematic approach helps break down complex problems into manageable parts, making problem-solving more confident and accurate.

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. Start by assigning letters to unknown quantities, such as $l$ for length or $w$ for width. Next, use the problem's information to write expressions or equations. For example, if the length is four times the width, write $l=\; 4w$. Then, use known formulas, like the perimeter of a rectangle $P=\; 2l+\; 2w$, to build an equation that relates all variables. This translation is crucial for solving the problem algebraically.

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

Substitution is a powerful technique when solving equations with multiple variables. If you have an equation relating two variables, such as $l=\; 4w$, you can substitute this expression into another equation involving both variables. For example, in the perimeter equation $P=\; 2l+\; 2w$, replace $l$ with $4w$ to get $P=\; 2(4w)\; +\; 2w$. This reduces the equation to one variable, making it easier to solve. Substitution simplifies complex problems by combining multiple equations into a single solvable equation.

Why is it important to check your solution in both the equation and the context of the problem?

Checking your solution in both the equation and the problem's context ensures accuracy and reasonableness. First, substitute your solution back into the original equation to verify that it satisfies the mathematical relationship. For example, if you found $w=\; 50$ and $l=\; 200$, plug these into $P=\; 2l+\; 2w$ to confirm the perimeter equals 500. Second, assess whether the solution makes sense in real life. If the perimeter is 500 yards, a length of 600 yards would be illogical. This two-step check helps catch errors and confirms that your answer is both mathematically correct and practically reasonable.

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

To build an equation for the perimeter of a rectangle, first identify the length and width variables, often denoted as $l$ and $w$. The perimeter is the total distance around the rectangle, which includes two lengths and two widths. Mathematically, this is expressed as $P=\; 2l+\; 2w$. In word problems, use this formula to relate the perimeter to the unknown dimensions. If additional relationships exist, such as the length being a multiple of the width, incorporate those to form a solvable equation.