Problem 30 - Algebra 1

Question

Use linear combinations to solve the system of linear equations. $$\begin{aligned}&0.1 g-h+4.3=0\\\&3.6=-0.2 g+h\end{aligned}$$

Step-by-Step Solution

Answer

The solution to the system of equations is \( g=5.76, h=1.16 \).

Step 1: Multiply by Appropriate Factors

In order to eliminate one of the variables, multiply the two equations by factors that will make the coefficients of one of the variables the same (ignoring the signs) in both equations. Multiply the first equation by 2 and the second by 10. The transformed system of equations is now: \[ \begin{aligned} &0.2 g-2 h+8.6=0, \&-2 g+10 h-36=0. \end{aligned} \]

Step 2: Add the Equations

Now, proceed to add the two equations together, \(0.2 g-2 h+ 8.6 + -2 g+10 h - 36 =0\), which simplifies to \(-1.8 g+8 h = -27.4.\)

Step 3: Solve for One Variable

Rearrange this equation to solve for g: \(g = \frac{8 h+27.4}{1.8}\).

Step 4: Substitute g into one of the original equations

Substitute the value of g from step 3 into the second equation from the original system. This will give one equation with only h: \(3.6 = -0.2 \frac{8 h+27.4}{1.8}+ h\). Which simplifies and can be solved to give: \(h = 1.16\).

Step 5: Solve for Second Variable

Substitute \(h = 1.16\) into the equation for g from step 3: \(g = \frac{8 * 1.16+27.4}{1.8}\). The solution is \(g = 5.76\).

Key Concepts

Understanding Systems of Linear Equations

Systems of linear equations consist of two or more linear equations that share the same set of variables. The main goal is to find values for these variables that make all the equations true simultaneously. In simple terms, you're looking for a common solution for all equations in the system. For example, when you have two variables, this system can be represented graphically as two lines on a coordinate plane.

When these lines intersect, the point of intersection represents the solutions where both equations are satisfied. If the lines are parallel, there is no solution because they never meet. However, if the lines overlap completely, there are infinitely many solutions because any point on the line will satisfy both equations.

This exercise deals with a classic system of two linear equations:

• Equation 1: \(0.1g - h + 4.3 = 0\)
• Equation 2: \(3.6 = -0.2g + h\)

Solving such a system involves finding the values for the variables \(g\) and \(h\) that satisfy both equations.

Elimination Method Explained

The elimination method is a technique used to solve a system of linear equations. It involves manipulating the equations to eliminate one variable, making it a single-variable equation that can be solved easily. The core idea is to add or subtract equations so that one of the variables cancels out.

In the provided exercise, elimination is applied by first transforming the coefficients of variables so they are equal, allowing their elimination through addition or subtraction:

• Multiply the first equation by 2: \(0.2g - 2h + 8.6 = 0\)
• Multiply the second equation by 10: \(-2g + 10h - 36 = 0\)

When adding these two transformed equations, \(-1.8g + 8h = -27.4\), the variable \(g\) is eliminated. The goal is to isolate one variable, say \(h\), and then solve for that variable. Once you determine \(h\), back-substitution into one of the original equations enables solving for \(g\).

This method is efficient when you want to remove a variable without manipulating the entire system overly.

Applying the Substitution Method

The substitution method is another effective strategy for solving systems of linear equations. This method involves solving one equation for one variable and then substituting that solution into the other equation. This approach turns the original system into a series of simpler calculations.

In our exercise, once elimination gives us a single equation in terms of \(h\), we solve it to get \(h = 1.16\). This calculated value of \(h\) is then substituted back into one of the original equations to find the value of \(g\).

Here’s how substitution works:

• Substitute \(h = 1.16\) into the expression for \(g\): \[g = \frac{8(1.16) + 27.4}{1.8}\]
• This results in \(g = 5.76\).

This method is particularly useful when one of the variables is already isolated in one equation, or can be easily isolated, allowing for quick substitution into the other equation. It's like solving a puzzle, one piece at a time, until you complete the entire picture.