Problem 11 - Algebra 2

Question

Solve each system of equations. \(5 x+2 y=4\) \(3 x+4 y+2 z=6\) \(7 x+3 y+4 z=29\)

Step-by-Step Solution

Answer

x = 2, y = -3, z = 6

Step 1: Identify the system of equations

The given system of equations is:1. \(5x + 2y = 4\)2. \(3x + 4y + 2z = 6\)3. \(7x + 3y + 4z = 29\)We need to find the values of \(x\), \(y\), and \(z\) that satisfy all three equations simultaneously.

Step 2: Express one variable in terms of others

From the first equation \(5x + 2y = 4\), solve for \(x\) in terms of \(y\):\[x = \frac{4 - 2y}{5}\]

Step 3: Substitute expression into other equations

Substitute \(x = \frac{4 - 2y}{5}\) into the second and third equations:For the second equation:\[3\left(\frac{4 - 2y}{5}\right) + 4y + 2z = 6\]Rewrite it:\[\frac{12 - 6y}{5} + 4y + 2z = 6\]Multiply everything by 5 to eliminate fractions:\[12 - 6y + 20y + 10z = 30\]Simplify it:\[14y + 10z = 18\]\For the third equation:\[7\left(\frac{4 - 2y}{5}\right) + 3y + 4z = 29\]Rewrite it:\[\frac{28 - 14y}{5} + 3y + 4z = 29\]Multiply everything by 5:\[28 - 14y + 15y + 20z = 145\]Simplify it:\[y + 20z = 117\]

Step 4: Solve the new system of equations

Now, solve the new system of two equations:1. \(14y + 10z = 18\)2. \(y + 20z = 117\)From the second equation, solve for \(y\):\[y = 117 - 20z\]Substitute back into the first equation:\[14(117 - 20z) + 10z = 18\]Calculate:\[1638 - 280z + 10z = 18\]Simplify:\[-270z = -1620\]Divide by -270:\[z = 6\]

Step 5: Find the value of y

Substitute \(z = 6\) back into \(y = 117 - 20z\):\[y = 117 - 20(6)\]Calculate:\[y = 117 - 120\]\[y = -3\]

Step 6: Find the value of x

Substitute \(y = -3\) back into \(x = \frac{4 - 2y}{5}\):\[x = \frac{4 - 2(-3)}{5}\]Calculate:\[x = \frac{4 + 6}{5}\]\[x = \frac{10}{5}\]\[x = 2\]

Step 7: Verify the solution

Verify that \(x = 2\), \(y = -3\), \(z = 6\) satisfy all original equations:- For the first equation: \(5(2) + 2(-3) = 10 - 6 = 4\) (True)- For the second equation: \(3(2) + 4(-3) + 2(6) = 6 - 12 + 12 = 6\) (True)- For the third equation: \(7(2) + 3(-3) + 4(6) = 14 - 9 + 24 = 29\) (True)

Step 8: Conclusion

The values \(x = 2\), \(y = -3\), \(z = 6\) solve the system of equations and satisfy all of the original equations.

Key Concepts

Linear Equations

Linear equations are mathematical expressions that represent lines when graphed on a coordinate plane. They consist of variables multiplied by constants and added together. In this context, we have a set of linear equations in three variables: \(x\), \(y\), and \(z\). Each equation has variables with different coefficients:

• \(5x + 2y = 4\)
• \(3x + 4y + 2z = 6\)
• \(7x + 3y + 4z = 29\)

The primary goal is to find values of \(x\), \(y\), and \(z\) that satisfy all these conditions simultaneously. Solutions to linear equations are often found using methods like substitution or elimination.

Substitution Method

The substitution method is a technique for solving a system of equations where you solve one equation for one variable and then substitute that expression into the other equation(s). In our exercise, we started with the first equation \(5x + 2y = 4\) and solved for \(x\) in terms of \(y\):

• \(x = \frac{4 - 2y}{5}\)

This expression was then substituted into the remaining equations. The substitution allows us to reduce the number of variables in the equation, making it easier to solve for the others. It effectively transforms the problem into one with fewer unknowns. The result of substituting \(x\) in terms of \(y\) helped us establish two resonable two-variable equations that we could solve more straightforwardly.

Solution Verification

After calculating potential solutions, it is crucial to verify them by substituting back into the original set of equations. In our exercise, the values found were \(x = 2\), \(y = -3\), and \(z = 6\). Verification involves checking:

• First equation: \(5(2) + 2(-3) = 10 - 6 = 4\), which is true.
• Second equation: \(3(2) + 4(-3) + 2(6) = 6 - 12 + 12 = 6\), also true.
• Third equation: \(7(2) + 3(-3) + 4(6) = 14 - 9 + 24 = 29\), again true.

Each calculation aligns perfectly with the given equations, confirming that our solutions are correct. Solution verification is a crucial final step, as it ensures that the derived values accurately satisfy all the original equations.

Algebraic Manipulation

Algebraic manipulation is a fundamental skill used to rearrange equations and expressions, making them simpler to evaluate. In this exercise, we employed several algebraic techniques:

• Solving for one variable in terms of others in a single equation, such as obtaining \(x = \frac{4 - 2y}{5}\) from \(5x + 2y = 4\).
• Substituting expressions from one equation into others to eliminate variables, reducing the complexity of the system.
• Ensuring balance by performing the same operations on both sides of the equation, evidenced by multiplying entire equations to clear denominators.
• Simplifying resulting equations to make them more straightforward to solve, such as turning \(14y + 10z = 18\) into a form that is easier to analyze.

These manipulations are essential for finding the solutions systematically and efficiently, ensuring clarity and direction in solving systems of equations.