Problem 40 - College Algebra

Question

Solve each system by any method. $$ \begin{array}{r} 0.1 x+0.2 y=2 \\ 0.35 x-0.3 y=0 \end{array} $$

Step-by-Step Solution

Answer

The solution is \(x = 6\) and \(y = 7\).

Step 1: Solve for One Variable in One Equation

We start by solving the second equation for one of the variables. The second equation is:\[ 0.35x - 0.3y = 0 \]Let's solve for \(x\) in terms of \(y\). Rearrange the equation:\[ 0.35x = 0.3y \]Now, solve for \(x\):\[ x = \frac{0.3y}{0.35} \]

Step 2: Simplify the Expression

Continue simplifying the equation for \(x\):\[ x = \frac{3y}{3.5} \]To simplify further:\[ x = \frac{6y}{7} \]

Step 3: Substitute into the First Equation

Now substitute the expression for \(x\) into the first equation:\[ 0.1\left(\frac{6y}{7}\right) + 0.2y = 2 \]Simplify this:\[ \frac{0.6y}{7} + 0.2y = 2 \]

Step 4: Clear the Fraction

To eliminate the fraction, multiply every term by 7:\[ 0.6y + 1.4y = 14 \]

Step 5: Combine Like Terms

Combine the terms on the left side of the equation:\[ 2y = 14 \]

Step 6: Solve for y

Divide both sides by 2 to solve for \(y\):\[ y = 7 \]

Step 7: Substitute back to find x

Use the expression for \(x\) from Step 2 to find \(x\). We have:\[ x = \frac{6(7)}{7} \]Calculate:\[ x = 6 \]

Step 8: Check the Solution

To ensure the solution is correct, substitute \(x = 6\) and \(y = 7\) back into the original equations:For the first equation:\[ 0.1(6) + 0.2(7) = 0.6 + 1.4 = 2 \]This satisfies the first equation.For the second equation:\[ 0.35(6) - 0.3(7) = 2.1 - 2.1 = 0 \]This satisfies the second equation as well.

Key Concepts

Substitution Method

The substitution method is a technique used to solve systems of equations by expressing one variable in terms of the other. This is accomplished by solving one of the equations in the system for one variable and then substituting this expression into the other equation. This method is particularly useful when one of the equations is easily solvable for a single variable.

In our original exercise, the equations given are:

• Equation 1: \(0.1x + 0.2y = 2\)
• Equation 2: \(0.35x - 0.3y = 0\)

We choose the second equation to solve for \(x\) because it can be easily manipulated to isolate \(x\) by moving terms around. This step helps simplify calculations in the subsequent equation, ultimately leading to a more straightforward problem-solving process.

Solving Equations

Solving equations involves finding the values of variables that satisfy given equations. For linear equations, this generally means manipulating the equations until you find clear solutions for each variable.

Once the substitution method is applied and one variable is expressed in terms of the other, it becomes a matter of substituting this expression into the second equation in the system. In our exercise, we've expressed \(x\) as \(\frac{6y}{7}\) from equation two. We then substitute this expression into the first equation, \(0.1x + 0.2y = 2\), resulting in one equation with only one variable.

After substitution, what remains is a typical problem of simplifying and solving a single linear equation, which can involve combining like terms or clearing fractions.

Algebraic Manipulation

Algebraic manipulation refers to the ability to transform equations through various algebraic operations such as addition, subtraction, multiplication, and division.

In the solved exercise, we had to simplify the expression for \(x\). Initially, it was expressed as \(\frac{0.3y}{0.35}\), which was simplified further to \(\frac{6y}{7}\). This simplification is crucial to avoid complications later in calculations. In dealing with fractions, multiplying each term by a common denominator is a technique used here. This action was demonstrated when eliminating the fraction by multiplying through by 7, giving us a clearer equation to solve.

Effective algebraic manipulation enables solving one-variable equations more efficiently, ultimately leading to discovering the solution, \(y = 7\), and subsequently \(x = 6\). Such techniques reinforce understanding of solving equations and working with systems of linear equations seamlessly.