Problem 90 - Elementary Algebra

Question

In the following exercises, solve the systems of equations by substitution. $$ \left\\{\begin{array}{l} 4 x+y=10 \\ x-2 y=-20 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system is \( x = 0 \) and \( y = 10 \).

Step 1: Solve one equation for one variable

First, solve the second equation for x: Given: ewline \[ x - 2y = -20 \] Add 2y to both sides: \[ x = 2y - 20 \]

Step 2: Substitute into the other equation

Next, substitute \( x = 2y - 20 \) into the first equation: Given: ewline \[ 4x + y = 10 \] Substitute \( x = 2y - 20 \):\[ 4(2y - 20) + y = 10 \]

Step 3: Simplify and solve for y

Simplify the equation: \[ 8y - 80 + y = 10 \] Combine like terms: \[ 9y - 80 = 10 \] Add 80 to both sides: \[ 9y = 90 \] Divide both sides by 9: \[ y = 10 \]

Step 4: Solve for x

Substitute the value of y back into the equation \( x = 2y - 20 \): \[ x = 2(10) - 20 \] Simplify: \[ x = 20 - 20 \] \[ x = 0 \]

Step 5: Verify the solution

Substitute \( x = 0 \) and \( y = 10 \) into the original equations to ensure they hold true: First equation: \[ 4(0) + 10 = 10 \] True. Second equation: \[ 0 - 2(10) = -20 \] True. Thus, the solution \( x = 0 \) and \( y = 10 \) is correct.

Key Concepts

Linear Equations

Linear equations are mathematical expressions that represent straight lines when graphed. They typically take the form \ ax + by = c \, where \ a \ and \ b \ are coefficients, \ x \ and \ y \ are variables, and \ c \ is a constant. These equations can be part of a system of linear equations, like in the given problem. The goal here is to find values for \ x \ and \ y \ that satisfy both equations simultaneously.
In the given exercise, we work with two linear equations:
\[ 4x + y = 10 \]
and
\[ x - 2y = -20 \].

This system represents two lines on a graph, and solving it means finding the point where these two lines intersect.

Algebraic Substitution

Algebraic substitution is a useful method for solving systems of equations. The idea is to solve one equation for one of the variables and then substitute this expression into the other equation.
Here's a quick step-by-step guide:

• Solve one of the equations for one variable.
• Substitute that expression into the other equation.
• Simplify and solve for the remaining variable.

In this problem, we first solve the second equation for \ x \:
\[ x - 2y = -20 \]
Adding \ 2y \ to both sides, we get:
\[ x = 2y - 20 \]
Then, we substitute this expression for \ x \ into the first equation:
\[ 4(2y - 20) + y = 10 \].

This allows us to solve for \ y \, simplifying our system to one equation with one unknown.

Variables and Constants

Understanding variables and constants is crucial in algebra. Variables are symbols that represent unknown values and can change, while constants are fixed values that do not change.
In our problem, \ x \ and \ y \ are variables, and the numbers 4, 1 (implicit in front of y), -20, and 10 are constants.
When we isolate \ y \ in our equations:

• First, we rewrite \ x \ in terms of \ y \: \[ x = 2y - 20 \].
• Then, we substitute this into the first equation, giving us constants \ 8 \ and -80 alongside variable \ y \: \[ 8y - 80 + y = 10 \].

Combining like terms and solving for \ y \ leads us to:
\[ y = 10 \].

With \ y \ known, substituting back to find \ x \ involves straightforward arithmetic with constants:
\[ x = 2(10) - 20 \], giving us \ x = 0 \.

Solution Verification

After solving for the variables, it's essential to verify that these solutions satisfy the original equations. This step ensures there's no mistake in the calculation process.
Given our solutions \ x = 0 \ and \ y = 10 \:

• Substitute \ x = 0 \ and \ y = 10 \ into the first equation:
  \[ 4(0) + 10 = 10 \] which holds true.
• Substitute the same values into the second equation:
  \[ 0 - 2(10) = -20 \] which also holds true.

These verifications confirm our solution: \ x = 0 \ and \ y = 10 \ is correct and satisfies both original equations. Always verify your solutions to confirm accuracy and correctness in solving systems of equations.