Problem 40 - Algebra 1

Question

Solve the linear system. $$ \begin{aligned} &3 x-5 y=3\\\ &9 x-20 y=6 \end{aligned} $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 0, y = -\frac{3}{5}\).

Step 1: Multiply the first equation by a suitable coefficient

The need to eliminate one variable in these equations. Therefore, multiply the first equation by 3 to make the \(x\) coefficient equal in both equations. This yields a new system of equations: \[\begin{aligned} &9x - 15y = 9\ &9x - 20y = 6 \end{aligned} \]

Step 2: Subtract the first equation from the second

Subtract the first new equation from the second, this will cancel out the \(x\) variable. \[ \begin{aligned}&9x - 9x - 20y + 15y = 6 - 9 \&0x + 5y = -3 \&5y = -3\end{aligned} \]

Step 3: Solve for the remaining variable

Simplify the equation to find the value of the remaining variable \(y\).\[ 5y = -3 \y = -\frac{3}{5}\]

Step 4: Substitute the value of \(y\) into the original equation

Substitute the value of \(y\) into the original equation to solve for \(x\).\[ 3x - 5(-\frac{3}{5}) = 3 \3x = -3 + 3 \3x = 0 \x = 0 \]

Key Concepts

Linear Equations

Understanding linear equations is fundamental to solving systems of equations. A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. Linear equations can be represented in the form of
\( ax + by = c \), where \(a\) and \(b\) are coefficients, \(x\) and \(y\) are variables, and \(c\) is a constant.

In the context of the given exercise:

1. \(3x - 5y = 3\) is the first linear equation.
2. \(9x - 20y = 6\) is the second linear equation.

The solutions to these equations are the values of \(x\) and \(y\) that make both equations true at the same time. A solution to a linear system is where the graphs of the equations intersect on a coordinate plane, which represents a unique point (\(x, y\)) unless the system has no solution or an infinite number of solutions.

Variable Elimination

Variable elimination is a technique used to find the solution to a system of linear equations by removing one variable to simplify the solution process. It involves using the addition or subtraction of the equations after manipulating them, if necessary, to align the coefficients of one variable.

Following the exercise presented, step 1 involves adjusting the first equation by multiplying it by 3 to align the \(x\) coefficients, resulting in a new pair of equations with the same \(x\) coefficient. This careful choice ensures that when one equation is subtracted from the other, as seen in step 2, the \(x\) terms cancel out, leaving an equation with only one variable:
\[5y = -3\].

This simplification is the crux of variable elimination and it strategically reduces a two-variable problem into a single-variable problem, making it straightforward to solve for the remaining variable.

Substitution Method

The substitution method is another approach to solving a system of linear equations, where one variable is expressed in terms of another, and this expression is substituted into the other equation. It is particularly handy when you can easily isolate one variable in one of the equations.

In this exercise, after elimination, we found that \(y = -\frac{3}{5}\). With this value, we can use the substitution method to find the value of \(x\). We substitute \(y\) in one of the original equations and solve for \(x\). The substitution in step 4 uses the first equation:\[3x - 5\left(-\frac{3}{5}\right) = 3\].

Simplifying further gives us \(3x = 0\), from which we get \(x = 0\). This substitution process provides a concrete value for the second variable, leading to the complete solution of the system.