Problem 45 - Algebra A Combined Function

Question

Solve each system by the substitution method. First simplify each equation by combining like terms. $$ \left\\{\begin{array}{l} -5 y+6 y=3 x+2(x-5)-3 x+5 \\ 4(x+y)-x+y=-12 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \((x, y) = (1, -3)\).

Step 1: Simplify the First Equation

The first equation is \(-5y + 6y = 3x + 2(x - 5) - 3x + 5\). Start by simplifying the left side: \(-5y + 6y = y\). Now simplify the right side: \(3x + 2(x - 5) - 3x + 5\) becomes \(3x + 2x - 10 - 3x + 5\) or equivalently \(2x - 5\). Now the equation is \(y = 2x - 5\).

Step 2: Simplify the Second Equation

The second equation is \(4(x + y) - x + y = -12\). Distribute the 4 and simplify: \(4x + 4y - x + y = -12\). Combine like terms: \(3x + 5y = -12\).

Step 3: Substitute the Expression for y

From Step 1, we have \(y = 2x - 5\). Substitute \(2x - 5\) for \(y\) in the simplified second equation \(3x + 5y = -12\). This gives us \(3x + 5(2x - 5) = -12\).

Step 4: Solve for x

Expand and simplify the equation from Step 3: \(3x + 10x - 25 = -12\). Combine like terms: \(13x - 25 = -12\). Add 25 to both sides: \(13x = 13\). Divide by 13 to solve for \(x\): \(x = 1\).

Step 5: Solve for y

Use the expression for \(y\) from Step 1: \(y = 2x - 5\). Substitute \(x = 1\) into this equation: \(y = 2(1) - 5 = 2 - 5 = -3\).

Step 6: Verify the Solution

Substitute \((x, y) = (1, -3)\) back into the original equations to verify the solution. For the first equation: \(-5(-3) + 6(-3) = 3(1) + 2(1 - 5) - 3(1) + 5\) simplifies to \(15 - 18 = 3 - 8 - 3 + 5\), which is \(-3 = -3\). For the second equation: \(4(1 + (-3)) - 1 + (-3) = -12\) simplifies to \(-8 = -12\), confirming the solution is correct.

Key Concepts

System of Equations

A system of equations is a set of two or more equations with the same variables. The solution to the system is the point where all these equations intersect in a coordinate plane.
To tackle these systems, we need to find values of the variables that make all the equations true simultaneously. In this exercise, we are tasked with solving a system with two equations involving two variables, \(x\) and \(y\). By simplifying and equating these equations, we can isolate the variables. This approach allows us to find a solution that satisfies both equations. Beyond just solving for the variables independently, systems of equations show how these variables relate to one another across different contexts or constraints in mathematics.

Solving Equations

Solving equations is about finding the value of the unknown variable that makes the equation true. In the substitution method, this process involves several clear steps:

• First, simplify each equation by combining like terms if necessary.
• Next, express one variable in terms of the other using one of the equations.
• Then, substitute this expression into the other equation. This substitution allows you to have one equation with one variable.
• Solve for the remaining variable.
• Finally, use the solved value to find the other variable.

These steps were demonstrated in the exercise where we first solved for \(y\) in terms of \(x\), substituted this into the second equation, and then solved for both variables. This method is effective since it reduces a multivariable problem into a single-variable one, simplifying the process.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations. They are used to represent real-world problems mathematically. In the context of systems of equations, algebraic expressions help us form the equations themselves.
For instance, the exercise equations were filled with algebraic expressions such as \(-5y + 6y\) and \(3x + 2(x-5) - 3x + 5\). To solve them, we engage in:

• Combining like terms, which simplifies the expression.
• Distributing operations, especially when variables are inside parentheses.
• Re-arranging terms to isolate specific variables.

These actions are fundamental in translating and solving real-world scenarios algebraically, as they form the backbone of how we simplify and interpret systems of equations.