Problem 39 - Algebra 2

Question

Solve each system of equations by using either substitution or elimination. \(2 f+2 g=18\) \(\frac{1}{6} f+\frac{1}{3} g=1\)

Step-by-Step Solution

Answer

The solution is \(f = 12\) and \(g = -3\).

Step 1: Simplify the second equation

First, let's eliminate the fractions in the second equation to make it easier to work with. Multiply the entire equation by 6 to clear the denominators:\[6 \left( \frac{1}{6}f + \frac{1}{3}g \right) = 6 \times 1\]This simplifies to:\[f + 2g = 6\]

Step 2: Set up the system of equations

Now, you have the following system of equations:\[\begin{align*}2f + 2g &= 18 \f + 2g &= 6\end{align*}\]

Step 3: Use elimination to solve for one variable

Subtract the second equation from the first equation to eliminate the variable \(g\):\[\begin{align*}(2f + 2g) - (f + 2g) &= 18 - 6 \f &= 12\end{align*}\]So, we have found that \(f = 12\).

Step 4: Substitute the found value into one of the original equations

Substitute \(f = 12\) back into the simplified equation from Step 1:\[12 + 2g = 6\]Solve for \(g\):\[2g = 6 - 12 \2g = -6 \g = -3\]

Step 5: Write the solution

The solution to the system of equations is \(f = 12\) and \(g = -3\).

Key Concepts

Substitution Method

The substitution method is a strategy for solving systems of linear equations. It involves solving one of the equations for one variable, and then substituting this expression into the other equation. This method is particularly useful when one equation is easily solvable for one of the variables.

For example, if you have the system of equations:

• \(x + y = 10\)
• \(2x - 3y = 5\)

you can solve the first equation for \(x\) in terms of \(y\):\[x = 10 - y\]Then, substitute \(10 - y\) for \(x\) in the second equation. This will allow you to solve for \(y\). Finally, use the value of \(y\) to find \(x\).

Substitution is advantageous because it reduces the system of equations to a single equation with one variable, making it simpler to solve.

Elimination Method

The elimination method is another powerful technique for solving systems of equations. This method involves adding or subtracting equations to eliminate one of the variables, allowing you to solve for the other variable directly.

Here's how it works with a simple system:

• \(3x + 2y = 14\)
• \(x - 2y = 2\)

To eliminate \(y\), you can add both equations together:\[(3x + 2y) + (x - 2y) = 14 + 2\]which simplifies to \(4x = 16\), allowing you to solve for \(x\). Then, substitute \(x\) back into one of the original equations to find \(y\).

The elimination method is particularly handy when the coefficients of one of the variables are the same or can easily be made the same, as it allows for a straightforward reduction of the system to one equation.

Solving Equations

Solving equations is a fundamental skill in mathematics that allows you to find the value of unknown variables. The process typically involves isolating the variable by performing inverse operations such as addition, subtraction, multiplication, or division. This requires maintaining the balance of the equation by performing the same operation on both sides.

For instance, in the equation \(2x + 3 = 11\), you can isolate \(x\) by first subtracting 3 from both sides:\[2x = 8\]Next, divide both sides by 2 to solve for \(x\):\[x = 4\]

These steps demonstrate the importance of keeping the equation balanced while systematically isolating the variable, a critical concept in solving equations.

Fractions in Equations

Fractions can make solving equations seem more complicated, but with the right techniques, they can be managed effectively. The key is to eliminate fractions at the start, which simplifies the process significantly.

Consider the equation:\[\frac{1}{2}x + \frac{1}{3} = \frac{5}{6}\]To eliminate the fractions, multiply every term by the least common denominator, which in this case is 6:\[6 \times \frac{1}{2}x + 6 \times \frac{1}{3} = 6 \times \frac{5}{6}\]This simplifies to:\[3x + 2 = 5\]Now, you have an equation without fractions, which is much easier to handle.

By clearing fractions early on, you not only simplify the equation but also reduce the risk of errors, making it easier and faster to find the solution.