Problem 31 - Algebra 1: Concepts and Skills

Question

Use linear combinations to solve the linear system. Then check your solution. \(x+2 y=5\) \(5 x-y=3\)

Step-by-Step Solution

Answer

The solutions to the system of equations are approximately \(x = 2.07\) and \(y = 1.465\).

Step 1: Multiply Equations

To create a linear combination that will allow us to eliminate one variable (either \(x\) or \(y\)), it is necessary to multiply the equations by suitable numbers. Here, if we multiply the first equation with 5 and the second equation by 2, then we would be able to eliminate the variable \(y\). The equations become: \n\(5x + 10y = 25\) \n\(10x - 2y = 6\)

Step 2: Add the Equations

Now, add the two equations. This operation eliminates variable \(y\). This yields a new equation: \(15x = 31\)

Step 3: Solve for \(x\)

Divide by 15 on both sides to solve for \(x\): \(x = 31 / 15\) or approximately \(x = 2.07\) (rounded to two decimal places)

Step 4: Substitute the value of \(x\) into one of original equations

We substitute \(x = 2.07\) into the first original equation (This step could be done with the second equation instead). This yields the equation: \(2.07 + 2y = 5\)

Step 5: Solve for \(y\)

Solve the equation to find \(y\). This leads to \(y = (5 - 2.07) / 2 \) or approximately \(y = 1.465\) (rounded to three decimal places)

Step 6: Check the obtained solutions

Finally, check whether \(x = 2.07\) and \(y = 1.465\) are the correct solutions by substituting these values into the initial system of equations.

Key Concepts

Linear Combinations

In algebra, the technique of linear combinations involves adding or subtracting equations to eliminate one of the variables, making it easier to solve a system of equations. This method turns a system of multiple linear equations into simpler, solvable equations. For example, when confronted with a procedure like in our exercise, we multiply each equation by a certain number to prepare them for combination.

Once prepared, adding or subtracting the equations results in one variable canceling itself out. In the case of our exercise, the multiplication of the first equation by 5 and the second by 2 set the stage for eliminating the variable y, resulting in a single linear equation that can be easily solved for x, dramatically simplifying the problem.

System of Equations

A system of equations is a set of two or more equations with the same variables. Systems of equations can have one solution, no solution, or infinitely many solutions. A solution to a system of linear equations is a set of values for the variables that make all the equations true simultaneously. In this case, the system comprises two equations with two variables, x and y. Solving these systems often involves finding the point where the two lines represented by the equations intersect on a graph.

Our provided exercise poses such a system, where we are looking for the intersection point of two lines, or the specific values of x and y that satisfy both equations.

Substitution Method

The substitution method is another way to solve a system of equations by finding the value of one variable and then substituting this value into the other equation. First, you solve one of the equations for one of the variables, then replace this variable in the other equation with the found value.

Applying this method to our exercise, after finding the value of x, which is approximately 2.07, we substitute this value into one of the original equations to solve for y. This step-by-step approach helps to isolate each variable, making it easier to calculate the exact or approximate solution for each.

Elimination Method

The elimination method for solving a system of linear equations involves adding or subtracting the equations to cancel out one of the variables, which facilitates finding the solution for the remaining variable. This method often involves multiplying one or both equations by a number to align the coefficients before elimination can occur.

In our exercise, after preparing the equations through multiplication, adding them directly leads to eliminating y, making it possible to solve for x. This method is particularly efficient when variables are easily eliminated with straightforward addition or subtraction, especially in systems where variables have coefficients amenable to quick cancellation.