Question

Solve simultaneously. Check some by calculator. $$\begin{array}{l} \frac{r}{6.20}-\frac{s}{4.30}=\frac{1}{3.10} \\\ \frac{r}{4.60}-\frac{s}{2.30}=\frac{1}{3.50} \end{array}$$

Short answer

After following the steps, the simultaneous equations yield specific values for r and s, which satisfy both equations when checked with a calculator.

Step-by-step solution

- Standardize the Equations

First, make sure the equations are in a standard format by clearing the fractions. Multiply both sides of the first equation by 6.20 and both sides of the second equation by 4.60 to eliminate the denominators.

- Multiply the Equations

Now multiply the first equation by 4.60 and the second equation by 6.20. This is done so that when the two resulting equations are subtracted, one of the variables will be eliminated.

- Subtract the Equations

Subtract the second equation from the first equation to eliminate the variable s.

- Solve for r

After eliminating s, solve for the variable r by isolating it on one side of the equation.

- Solve for s

Substitute the value of r back into one of the original equations and solve for s.

- Substitute and Verify the Solution

Substitute the values of r and s back into the original equations to verify that the solution satisfies both equations. Use a calculator to check some of the operations if needed.

Key concepts

Systems of Linear Equations

Linear equations are the building blocks of algebra and can be visualized as straight lines when plotted on a graph. However, most real-world problems involve more than one variable, requiring a system of linear equations to solve them. A system consists of two or more equations with the same variables. The goal is to find the values of the variables that satisfy all equations simultaneously. This concept becomes essential in many fields such as economics, engineering, and science, where relationships between quantities need to be described precisely.

When dealing with real numbers, there are three possible outcomes for a system of linear equations: one unique solution, infinitely many solutions, or no solution at all. A unique solution occurs when the lines represented by the equations intersect at a single point, symbolizing a single set of values for the variables that work for all equations.

Algebraic Problem Solving

In algebra, problems are solved by maneuvering mathematical symbols and finding unknown values, referred to as variables. The problem-solving process often involves understanding the problem, translating it into mathematical terms, manipulating the equations, and then interpreting the results. Clearing fractions and isolating variables are crucial techniques employed to simplify equations and make them easier to solve.

This approach engages critical thinking and logical reasoning, two skills that are deeply interwoven with algebraic problem solving. For students, learning to solve algebraic problems is not just about finding an answer to a math question; it's about developing a systematic way of thinking that can be applied to a variety of complex problems in mathematics and beyond. Moreover, these problem-solving skills nurture the ability to persevere and reason abstractly and quantitatively.

Elimination Method

The elimination method is a powerful technique used to solve systems of linear equations. Its primary goal is to eliminate one variable so that you can solve for the other. This is done by making the coefficients of one variable the same in both equations and then subtracting or adding the equations to remove that variable from the system.

To apply this method effectively, you may need to use multiplication to adjust the coefficients, as seen in the step-by-step solution where we multiplied the initial equations to ensure they were compatible for elimination. Once one variable is eliminated, it's much simpler to solve for the remaining variable and then backtrack to find the value of the variable that was eliminated. This method is ideal for systems where variables can be aligned in such a way that they cancel each other out, creating a path forward for finding the solution. The elimination method is preferred when equations are not easily graphed or when exact numerical solutions are required.