Question

In order to find the tons \(x\) of steel containing \(5.25 \%\) nickel to be combined with another steel containing \(2.84 \%\) nickel to make 3.25 tons of steel containing 4.15\% nickel, we must solve the equation $$0.0525 x+0.0284(3.25-x)=0.0415(3.25)$$ Solve this equation,

Short answer

x \( \approx \) 1.79 tons

Step-by-step solution

Set up the equation

Let the amount of steel with 5.25% nickel be represented by x tons. The total steel mixture is 3.25 tons. This means the amount of steel with 2.84% nickel will be (3.25 - x) tons. The equation representing the nickel content in the final mixture is: $$0.0525x + 0.0284(3.25 - x) = 0.0415 \times 3.25$$

Distribute the 0.0284

Multiply 0.0284 with both 3.25 and -x to simplify the left side of the equation: $$0.0525x + (0.0284 \times 3.25) - (0.0284 \times x) = 0.0415 \times 3.25$$

Simplify the equation

After performing the multiplication, the equation becomes: $$0.0525x + 0.0923 - 0.0284x = 0.135375$$ Next, combine like terms by subtracting 0.0284x from 0.0525x to get: $$0.0241x + 0.0923 = 0.135375$$

Isolate the variable x

Subtract 0.0923 from both sides of the equation to isolate the terms with x: $$0.0241x = 0.135375 - 0.0923$$

Solve for x

Perform the subtraction on the right side and then divide both sides by the coefficient of x to find the value of x: $$0.0241x = 0.043075$$ $$x = \frac{0.043075}{0.0241}$$

Calculate the tons of steel

Dividing the two numbers gives the tons of steel with 5.25% nickel needed: $$x = 1.78755207$$ This is approximately 1.79 tons when rounded to two decimal places.

Key concepts

Mixture Problems

Mixture problems in algebra are a class of word problems that involve combining two or more substances with different properties to create a new substance with a specific desired property. The goal often includes finding out how much of each original substance is needed to obtain a certain amount of the final mixture.

These problems are approached by setting up algebraic equations that represent the quantities and characteristics (such as concentration) of the substances involved. For instance, if two types of steel with different nickel contents are mixed, we would use the nickel percentage of each type of steel to form our equation. It's important to label your variables carefully and understand that the total quantity of the final mixture is the sum of the quantities of the individual components.

Establishing relationships between the components and the final mixture is crucial in mixture problems, and it's often where students can improve their skills. Start by identifying the key quantities, list out what is known and what needs to be found, and ensure the variables represent real-life quantities that make sense in the context of the problem.

Nickel Content Calculation

The nickel content calculation is essential in mixture problems that involve metals or alloys. The content is usually expressed as a percentage, representing the amount of nickel per unit mass of the metal or alloy. This calculation is done by multiplying the mass of the metal by the percentage of nickel it contains, converted into a decimal.

To improve clarity in such problems, it's wise to specify the mass and nickel content of each individual metal. For example, if we have two types of steel with different nickel percentages, we'd calculate the mass of nickel in each type separately before combining them. After setting up the initial equation that represents the overall nickel content of the mixture, it's also practical to ensure that all percentages are correctly converted into decimal form to avoid common mistakes.

Providing visual aids, like diagrams or tables, can help put the calculations into perspective. For students, keeping track of units and maintaining consistent formats among the values in the problem are key aspects that can prevent errors and improve comprehension.

Algebraic Equations

Algebraic equations are the foundation of solving mixture problems. They are mathematical statements that assert the equality of two expressions and usually contain one or more unknown variables. The resolution of these equations involves finding the value(s) of these variables that make the equation true.

In the context of mixture problems, algebraic equations are formed by relating the quantities and concentrations of components in the mixtures. For a well-solved problem, each step of the solution needs to be clear and logical, taking care to combine like terms and isolate variables properly.

To aid understanding, it's valuable to explain the purpose of each equation-solving step. For instance, when combining like terms, you are simplifying the equation; when isolating the variable, you're getting closer to the solution. Practical tips for students tackling these equations include double-checking their work at each step, cross-multiplying to verify solutions, and using substitution to check if the obtained solution satisfies the original equation.