Question

Solve the linear system. $$ \begin{aligned} &1.5 x-2.5 y=8.5\\\ &6 x+30 y=24 \end{aligned} $$

Short answer

The solution to the system of equations is \(x = 5.75\) and \(y = -0.25\).

Step-by-step solution

Simplify the Equations

The second equation could simplify by dividing all terms by 6. The simplified equations are, \n \(1.5x - 2.5y = 8.5\), \n \(x + 5y = 4\).

Use Elimination or Substitution

In this case, elimination can be a better approach. Subtract the second equation from the first to eliminate \(x\). \n So, \(1.5x - 2.5y - (x + 5y) = 8.5 - 4\), \n Which simplifies to \(0.5x - 7.5y = 4.5\). Now solve this equation for \(x\). \n Multiplying by 2 to make calculation easier yields, \(x - 15y = 9\). Add 15y to both sides to solve for \(x\), we get \(x = 9 + 15y\). This equation gives any x in terms of y.

Substitute x Into Second Equation

Replace \(x\) in the second equation with the expression we found: \(9 + 15y + 5y = 4\). Simplify to find \(y\): \(20y = -5\), then divide by 20, giving \(y = -0.25\).

Substitute y Into Equation for x

Now that we have \(y\), we can substitute it back into the equation we found for \(x\) in Step 2: \(x = 9 + 15*(-0.25) = 5.75\). So, the solution to the system of equations is \(x = 5.75\) and \(y = -0.25\).

Key concepts

Solving Linear Systems

Understanding how to solve linear systems is a fundamental skill in algebra. A linear system consists of two or more linear equations involving the same set of variables. The goal is to find a value for each variable that makes all the equations true simultaneously. There are multiple methods for solving such systems, with the elimination and substitution methods being two of the most common.

When it comes to solving linear systems, consistency and the nature of the solution are crucial. A system can have a single unique solution, infinitely many solutions, or no solution at all—the latter being referred to as inconsistent. Determining which kind of system you are dealing with can often be done by examining the structure of the equations or by beginning to solve them using one of the methods.

Elimination Method

The elimination method is particularly useful when you can easily cancel out one variable by adding or subtracting the equations. The idea is to multiply or divide each equation by suitable numbers so that when you perform the addition or subtraction, one of the variables cancels itself out.

Steps to Use Elimination

• Multiply (or divide) one or both of the equations by a number that creates a coefficient in front of one of the variables that is the negation of the coefficient in front of the same variable in the other equation.
• Add or subtract the equations to eliminate that variable, resulting in an equation with only one variable.
• Solve for the remaining variable.
• Substitute the value back into one of the original equations to find the value of the other variable.

Applying the elimination method correctly simplifies the process of finding the solution and helps to avoid errors that might occur due to complex substitutions.

Substitution Method

The substitution method works by solving one of the equations for one variable in terms of the others and then substituting this expression into the other equations. This method works well when one equation is already solved for one variable, or can be easily manipulated to have one variable isolated on one side of the equation.

Steps to Use Substitution

• Choose one equation and solve it for one variable.
• Substitute the expression you found into the other equation(s).
• Solve the new equation that now has just one variable.
• Use the found value to solve for the other variable in one of the original equations.

This method often involves fewer steps than elimination and can be more straightforward if the equations are already set up to substitute one into another easily.

Simplifying Equations

Simplifying equations is a pivotal step before applying any method to solve them. To simplify, we can combine like terms, eliminate fractions, and reduce any coefficients to their lowest terms. Simplifying makes the rest of the problem easier to solve, and can often give insight into which method of solution might be more efficient. Always look for opportunities to divide both sides by a coefficient, if possible, to obtain simpler whole number coefficients to work with.

By simplifying the equation, as in the provided exercise where the second equation is divided by 6, it becomes more manageable and the subsequent steps of utilizing the elimination or substitution method can be performed with less risk of arithmetic mistakes.