Problem 4 - Algebra and Trigonometry with Analytic Geometry

Question

\(\left\\{\begin{aligned} 4 x-y+3 z &=6 \\\\-8 x+3 y-5 z &=-6 \\ 5 x-4 y &=-9 \end{aligned}\right.\)

Step-by-Step Solution

Answer

The solution is \(x = 3\), \(y = 6\), \(z = 0\).

Step 1: Identify the Variables and Equations

We have three equations with three variables: \(x\), \(y\), and \(z\). The system of equations is:1. \(4x - y + 3z = 6\)2. \(-8x + 3y - 5z = -6\)3. \(5x - 4y = -9\)

Step 2: Solve for One Variable

We'll solve Equation 3 for \(x\). From equation 3: \[ 5x - 4y = -9 \]Rearrange to isolate \(x\):\[ 5x = 4y - 9 \]\[ x = \frac{4y - 9}{5} \]

Step 3: Substitute for x in Other Equations

Substitute \(x = \frac{4y - 9}{5}\) into Equations 1 and 2.For Equation 1:\[ 4\left(\frac{4y - 9}{5}\right) - y + 3z = 6 \]Multiplying out the fraction:\[ \frac{16y - 36}{5} - y + 3z = 6 \]Clear the fraction by multiplying the whole equation by 5:\[ 16y - 36 - 5y + 15z = 30 \]Simplifying gives: \[ 11y + 15z = 66 \] (Equation 4)For Equation 2:\[ -8\left(\frac{4y - 9}{5}\right) + 3y - 5z = -6 \]Multiply out:\[ -\frac{32y - 72}{5} + 3y - 5z = -6 \]Clear the fraction by multiplying the whole equation by 5:\[ -32y + 72 + 15y - 25z = -30 \]Simplifying gives: \[ -17y - 25z = -102 \] (Equation 5)

Step 4: Solve New System of Equations

Now, we solve the system of equations formed by Equations 4 and 5:1. \(11y + 15z = 66\)2. \(-17y - 25z = -102\)Multiplying Equation 4 by 17:\[ 187y + 255z = 1122 \]Multiplying Equation 5 by 11:\[ -187y - 275z = -1122 \]Adding these two equations cancels \(y\):\[ (187 - 187)y + (255 - 275)z = 1122 - 1122 \]\[ -20z = 0 \]Solve for \(z\):\[ z = 0 \]

Step 5: Solve for y Using z

Substitute \(z = 0\) back into Equation 4:\[ 11y + 15(0) = 66 \]\[ 11y = 66 \]\[ y = 6 \]

Step 6: Solve for x Using y

Now use \(y = 6\) in the expression for \(x\) from Step 2:\[ x = \frac{4(6) - 9}{5} \]\[ x = \frac{24 - 9}{5} \]\[ x = \frac{15}{5} \]\[ x = 3 \]

Step 7: Verify the Solution

To ensure our solutions \(x = 3\), \(y = 6\), and \(z = 0\) are correct, substitute them back into all original equations:1. \(4(3) - 6 + 3(0) = 6\) which simplifies to \(12 - 6 = 6\), correct.2. \(-8(3) + 3(6) - 5(0) = -6\) which simplifies to \(-24 + 18 = -6\), correct.3. \(5(3) - 4(6) = -9\) which simplifies to \(15 - 24 = -9\), correct.

Key Concepts

Substitution Method

The substitution method is a useful technique for solving systems of equations, particularly when one equation can be easily rearranged to isolate one of the variables. This involves substituting an expression for the isolated variable into the other equations.
Here's how it works:

• Identify which equation you can solve easily for one variable. In our problem, Equation 3, which is \(5x - 4y = -9\), is a convenient choice because it involves only two variables.
• Rearrange this equation to solve for one of the variables. We solved for \(x\), giving us \(x = \frac{4y - 9}{5}\).
• Substitute this expression for \(x\) into the other equations. This helps eliminate \(x\), allowing us to solve for the remaining variables.

By using substitution, you take a potentially complicated system and reduce it step by step, making it more manageable.

Linear Equations

Linear equations are mathematical expressions that define a straight line when graphed on a coordinate plane. They take the form \(ax + by + cz = d\), where \(a\), \(b\), \(c\), and \(d\) are constants. These equations can involve multiple variables but are linear because each variable is to the first power.
In a system of linear equations, like in our exercise:

• Each equation graphically represents a plane.
• The solution, if it exists, is the point(s) where these planes intersect.
• For three variables (\(x\), \(y\), \(z\)), solving means finding where all three planes intersect at a single point, a line, or being parallel (no solution in case of parallel).

Linear systems may have one solution, infinitely many solutions, or no solution, depending on the relationship between the equations.

Variable Isolation

Variable isolation is the technique of rearranging an equation to solve for one variable, leaving it on one side of the equation by itself. This process is crucial for using the substitution method or any other method aimed at solving equations.Key steps include:

• Rearranging terms to position the desired variable on one side. This involves using basic algebraic operations like addition, subtraction, multiplication, or division.
• In our example, we isolated \(x\) in Equation 3: \(5x - 4y = -9\) was rearranged to \(x = \frac{4y - 9}{5}\).
• This isolation simplifies substitution into other equations, thus reducing the problem to fewer equations with fewer unknowns.

Once a variable is isolated, it can be substituted back into other equations, methodically leading to the final solution.