Problem 29 - Algebra 1: Concepts and Skills

Question

Use linear combinations to solve the linear system. Then check your solution. \(g-10 h=43\) \(18=-g+5 h\)

Step-by-Step Solution

Answer

The solution to the linear system is \(g=-79\) and \(h=-12.2\)

Step 1: Set Equal Coefficients

The first step is to set the coefficients of one of the variables, say g, equal in both equations. Currently, its coefficients are 1 and -1. Multiplying the second equation by 2 makes the coefficient of g in the second equation -2, allowing it to be cancelled out when added to the first equation:\n\n\(2(-g + 5h) = 2(18)\)\n\nSimplify this to get the new second equation: \n\n\(-2g + 10h = 36\)

Step 2: Add the Equations

Add the two equations to cancel out g:\n\n\((g - 10h) + (-2g + 10h) = 43 + 36\)\n\nThis simplifies to:\n\n\(-g = 79\)

Step 3: Solve for g

The next step is to isolate g in the equation. As g is negative, multiply the equation by -1 to get: \n\n\(g = -79\)

Step 4: Substitute g in the second equation

Substitute \(g = -79\) into the second equation \(-g+5h=18\) to find the value of h:\n\n\(-(-79) + 5h = 18\)\n\nThis gives:\n\n\(79 + 5h = 18\)

Step 5: Solve for h

Subtract 79 from both sides to isolate the term with h:\n\n\(5h = 18 - 79\)\n\nThen divide both sides by 5 to solve for h:\n\n\(h = (18 - 79) / 5\)\n\nSo, \(h=-12.2\)

Step 6: Check the Solution

Lastly, substitute g=-79 and h=-12.2 in both original equations to validate the solution. The left-hand side of both equations should equal the right-hand side if the solution is correct.

Key Concepts

Understanding Linear Systems

When we talk about linear systems, we mean a collection of two or more linear equations involving the same set of variables. In this exercise, the equations given are:

• \( g - 10h = 43 \)
• \( 18 = -g + 5h \)

These represent two equations with two variables, \( g \) and \( h \). The goal is to find the values of these variables that satisfy both equations simultaneously. To solve such a system, one popular method is using linear combinations. This involves adding or subtracting equations after appropriate manipulation to eliminate one of the variables. Once one variable is eliminated, the system can simplify significantly, allowing us to solve for the remaining variable painstakingly. Then, that value can be plugged back into one of the original equations to find the value of the other variable. By understanding this concept, you can solve multiple equations in real-world applications, like economics or engineering, where many variables interact simultaneously.

Learning to solve linear systems using linear combinations is one stepping stone towards grasping more complex algebraic systems.

Solving Equations with Linear Combinations

Solving equations using linear combinations, also known as the addition or elimination method, is a great way to tackle linear systems. Here’s a simple breakdown:

• First, adjust the coefficients of the variables in the equations so they can cancel each other out. In our problem, we multiplied the second equation by 2 to set up cancellation for \( g \).
• When added together, one variable is eliminated. This simplifies the problem because you focus on finding one variable at a time.

Let's look at our case:

• We manipulated equations so that \( g \) was eliminated, leading us easily to solve the simplified equation.
• This method is adaptable, meaning if one variable is difficult to eliminate by addition, a small tweak or multiplication of the whole equation often solves it.

Now, the simplest equation gives us \( g = -79 \). Solving equations with methods like this is crucial for algebra. Adjusting and manipulating equations to bring about zero in one variable is a handy skill. It simplifies complexities into manageable tasks, a principle behind algebraic efficiency.

The Art of Algebraic Manipulation

Algebraic manipulation refers to the skill of rearranging and adjusting algebraic expressions to make them easier to solve. It's like solving a puzzle; moving parts might seem arbitrary at first, but each change brings you closer to the solution. In solving the given exercises:

• We multiplied terms to set the coefficients necessary for elimination. This helps in removing a variable from the equation temporarily.
• We isolated the variable by reversing operations – dividing or subtracting where needed to strengthen algebraic expressions.

Think of algebraic manipulation as your toolset:

• It helps you create equivalent expressions, allowing equations to showcase their hidden simplicity after manipulation.
• Such tactical rearrangement helps simplify complex problems, focusing on not what the equation initially says but on its core information.

In these steps, multiplying, adding, or even negating entire equations are the levers and pulleys that make the machinery of algebra work smoothly. Practicing these skills turns intricate problems into straightforward calculations.