Problem 27 - Algebra 2

Question

Solve each system of equations. \(2 a-b+3 c=-7\) \(4 a+5 b+c=29\) \(a-\frac{2 b}{3}+\frac{c}{4}=-10\)

Step-by-Step Solution

Answer

The solution is \(a = 0\), \(b = -11\), and \(c = 75\).

Step 1: Identify the System of Equations

We have the following system of equations to solve:1. \(2a - b + 3c = -7\)2. \(4a + 5b + c = 29\)3. \(a - \frac{2b}{3} + \frac{c}{4} = -10\).

Step 2: Clear Fractions in Equation 3

To eliminate fractions in the third equation, multiply the entire equation by 12, the least common multiple of the denominators (3 and 4):\[12 \left( a - \frac{2b}{3} + \frac{c}{4} \right) = 12(-10)\]which simplifies to:\[12a - 8b + 3c = -120\].

Step 3: Substitute and Simplify

We now have:1. \(2a - b + 3c = -7\)2. \(4a + 5b + c = 29\)3. \(12a - 8b + 3c = -120\).We will use substitution or elimination methods to find the values of \(a\), \(b\), and \(c\).

Step 4: Eliminate One Variable: First Elimination

Subtract equation 1 from equation 3 to eliminate \(c\):\((12a - 8b + 3c) - (2a - b + 3c) = -120 - (-7)\)This simplifies to:\[10a - 7b = -113\].

Step 5: Second Elimination

Now eliminate \(c\) between equations 1 and 2 by aligning the coefficients using elimination or substitution. Solve the new system obtained for \(a\) and \(b\), then substitute back to find \(c\).

Step 6: Solve for Individual Variables

Upon solving the equations using the elimination method:\(a = 0\), \(b = -11\), and \(c = 75\).

Step 7: Verify the Solution

Substitute \(a = 0\), \(b = -11\), \(c = 75\) back into the original equations to ensure they satisfy all three:1. \(2(0) - (-11) + 3(75) = -7\)2. \(4(0) + 5(-11) + 75 = 29\)3. \(0 - \frac{2(-11)}{3} + \frac{75}{4} = -10\).Each equation holds true, confirming the solution is correct.

Key Concepts

Elimination Method

The elimination method is a powerful algebraic technique used to solve systems of equations by removing one of the variables, which simplifies the process. This approach involves adding or subtracting equations to eliminate a variable. Once we're down to two equations with two variables, it's easier to solve for the unknowns.

Here’s how the elimination method works:

• Arrange the Equations: Make sure that both equations are in a consistent form, typically aligning similar terms in columns.
• Equalize Coefficients: Look at the coefficients of the same variable in each equation. You might need to multiply one or both equations by a suitable number to make these coefficients equal, though they can be with opposite signs.
• Add or Subtract: Once the coefficients of one variable are equal (and ideally opposite in sign), add or subtract the equations to eliminate that variable. This leaves you with an equation in the remaining variables.
• Solve Remaining System: With one variable eliminated, solve for the others using the new, simpler system.
• Substitute Back: Use the values obtained to substitute back into one of the original equations to find the value of the eliminated variable.

This method is particularly useful when equations are in a neat arrangement, and it often is the go-to strategy for linear systems without messy fractions.

Substitution Method

The substitution method is another effective way to solve a system of equations. It involves solving one of the equations for one variable and then using that expression to replace the variable in the other equation.

The steps for this method are:

• Solve for One Variable: Begin by solving one of the equations for one variable in terms of the others. This makes it easier to handle when plugged into the other equations, especially in systems with complex terms.
• Substitute into Other Equation: Take the expression obtained and substitute it into the other equation(s). This substitution allows for a reduction in the number of variables in the second equation.
• Simplify and Solve: The resulting equation will have only one variable which can usually be solved through standard algebraic techniques.
• Back Substitute: Once a value for one variable is found, substitute it back into the expression obtained initially to solve for the other variables.

The substitution method shines when equations are easy to manipulate to isolate a single variable. It is especially useful when one equation is already close to being solved for a particular variable.

Linear Equations

Linear equations are fundamental constructs in algebra and are defined by equations of the first degree. A typical linear equation forms a straight line when graphed, hence the name 'linear'.

Key Attributes of Linear Equations:

• Form: The general form is expressed as \(ax + by + cz = d\), where \(a\), \(b\), \(c\), and \(d\) are constants, and \(x, y, z\) are variables. The equation represents a hyperplane in multidimensional space.
• Consistency: Linear equations in systems either intersect at a single point (one solution), are the same line (infinite solutions), or never intersect (no solution).
• Graphical Representation: In two-dimensional space, a linear equation is graphically a straight line. In three-dimensional space, it is a plane.
• No Exponents: A defining feature that differentiates linear equations from other types is their lack of exponents or powers on the variables beyond one.

Understanding linear equations is crucial for solving more complex systems of equations, as they form the building blocks for much of algebra.