Problem 129 - Elementary Algebra

Question

In the following exercises, solve the systems of equations by substitution. $$ \left\\{\begin{array}{l} 3 x-y=-7 \\ 4 x+2 y=-6 \end{array}\right. $$

Step-by-Step Solution

Answer

x = -2, y = 1

Step 1: Solve one equation for one variable

Take the first equation, which is \(3x - y = -7\). Solve for \(y\) in terms of \(x\):\[ y = 3x + 7 \]

Step 2: Substitute into the second equation

Substitute \(y = 3x + 7\) into the second equation \(4x + 2y = -6\):\[ 4x + 2(3x + 7) = -6 \]

Step 3: Simplify the equation

Simplify the equation from step 2:\[ 4x + 6x + 14 = -6 \]\[ 10x + 14 = -6 \]

Step 4: Solve for \(x\)

Solve the simplified equation for \(x\):\[ 10x = -6 - 14 \]\[ 10x = -20 \]\[ x = -2 \]

Step 5: Substitute \(x\) back into the first equation for \(y\)

Substitute \(x = -2\) into the equation \(y = 3x + 7\):\[ y = 3(-2) + 7 \]\[ y = -6 + 7 \]\[ y = 1 \]

Step 6: Write the solution

Combine the values of \(x\) and \(y\) to write the solution of the system:\( x = -2 \), \( y = 1 \)

Key Concepts

substitution method

The substitution method is a technique used to solve systems of linear equations by substituting one equation into another. This method simplifies the system into a single equation with one variable.

Here’s a step-by-step approach:

• First, solve one of the equations for one variable in terms of the other variable. For example, if given the equation \(3x - y = -7\), solve for \(y\) to get \(y = 3x + 7\).
• Next, substitute this expression into the other equation. In our second equation, \(4x + 2y = -6\), we replace \(y\) with \(3x + 7\), forming \(4x + 2(3x + 7) = -6\).
• Simplify the resulting equation and solve for the single variable.
• Lastly, substitute the value obtained back into one of the original equations to solve for the second variable.

The substitution method is particularly useful when one of the equations is easily solved for one of the variables, making it a preferred choice for systems where variables are aligned in a straightforward manner.

linear equations

Linear equations are algebraic expressions that represent straight lines when graphed. They are written in the form \(ax + by = c\) where \(a\), \(b\), and \(c\) are constants.

Key points about linear equations:

• They involve only first-degree variables, which means no variable is raised to a power other than one.
• These equations can have one or more variables.
• The graph of a linear equation in two variables is always a straight line.
• The solutions to linear equations represent points of intersection of these lines when plotted on a coordinate plane.

In our exercise, we dealt with two linear equations: \(3x - y = -7\) and \(4x + 2y = -6\). By using methods such as substitution, we can find the point where these two lines intersect, representing the solution to the system of equations.

algebraic manipulation

Algebraic manipulation involves the use of algebraic operations to simplify and solve equations. It is fundamental in working through systems of equations in methods like substitution.

Here are some basic algebraic manipulations used in solving our system of equations:

• Rewriting equations: In step one, we re-arranged \(3x - y = -7\) to solve for \(y\) as \(y = 3x + 7\).
• Substitution: We replaced \(y\) in the second equation with \(3x + 7\), leading to a single-variable equation.
• Simplification: In steps two and three, by performing operations like distribution and combining like terms, we reduced the equation \(4x + 2(3x + 7) = -6\) to its simplest form \(10x + 14 = -6\).
• Solving linear equations: Finally, in steps four and five, we isolated the variable \(x\) and solved, then substituted back to find \(y\).

Mastery of algebraic manipulation is essential for solving more complex mathematical problems and understanding the core structures of algebra.