Solving 1/2x + 1/5y = 1 And 1/3x - 1/5y = 4 Elimination And Substitution Methods

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Hey guys! Today, we're diving into the exciting world of solving systems of equations. Specifically, we'll be tackling the system:

  1. 1/2x + 1/5y = 1
  2. 1/3x - 1/5y = 4

We're going to use two powerful methods: elimination and substitution. These are like the dynamic duo of equation-solving, and mastering them will seriously level up your math game. So, let's jump right in!

Method 1: Elimination – The Art of Cancellation

Elimination, as the name suggests, involves strategically eliminating one of the variables to solve for the other. It's like a mathematical magic trick where we make one variable disappear!

The key idea here is to manipulate the equations so that the coefficients of either x or y are opposites. When we add the equations together, one variable will cancel out, leaving us with a single equation in one variable.

Looking at our system:

  1. 1/2x + 1/5y = 1
  2. 1/3x - 1/5y = 4

We notice something awesome: the coefficients of y are already opposites! We have +1/5y in the first equation and -1/5y in the second. This means we're already halfway there!

Step-by-Step Elimination

  1. Add the Equations: Since the y terms have opposite coefficients, we can simply add the two equations together:

    (1/2x + 1/5y) + (1/3x - 1/5y) = 1 + 4

    This simplifies to:

    1/2x + 1/3x = 5

  2. Combine the x Terms: To add the x terms, we need a common denominator for the fractions 1/2 and 1/3. The least common denominator is 6. So, we rewrite the fractions:

    (3/6)x + (2/6)x = 5

    Combining the terms, we get:

    (5/6)x = 5

  3. Solve for x: To isolate x, we multiply both sides of the equation by the reciprocal of 5/6, which is 6/5:

    (6/5) * (5/6)x = 5 * (6/5)

    This simplifies to:

    x = 6

    Boom! We've found the value of x. That wasn't so bad, right?

  4. Substitute x Back In: Now that we know x = 6, we can substitute this value into either of the original equations to solve for y. Let's use the first equation:

    1/2x + 1/5y = 1

    Substituting x = 6, we get:

    1/2(6) + 1/5y = 1

    Simplifying:

    3 + 1/5y = 1

  5. Solve for y: Subtract 3 from both sides:

    1/5y = -2

    Multiply both sides by 5:

    y = -10

    And there you have it! We've found the value of y.

The Solution

Using the elimination method, we've determined that the solution to the system of equations is x = 6 and y = -10. We can write this as an ordered pair (6, -10).

Why Elimination Works

The beauty of elimination lies in its efficiency. When the coefficients line up nicely (or can be easily manipulated to line up), it's a super-fast way to solve for the variables. It's like a mathematical shortcut that saves us time and effort.

Method 2: Substitution – The Art of Isolating and Replacing

Substitution is another powerful technique for solving systems of equations. It involves isolating one variable in one equation and then substituting that expression into the other equation. Think of it as replacing one variable with an equivalent expression.

This method is particularly useful when one of the equations is already solved for one variable or can be easily solved for one variable. It's like finding a key piece of information and using it to unlock the rest of the puzzle.

Let's revisit our system:

  1. 1/2x + 1/5y = 1
  2. 1/3x - 1/5y = 4

Step-by-Step Substitution

  1. Choose an Equation and Isolate a Variable: We need to pick one of the equations and solve it for either x or y. Let's choose the first equation and solve for y (we could also solve for x, but solving for y seems slightly easier in this case):

    1/2x + 1/5y = 1

    Subtract 1/2x from both sides:

    1/5y = 1 - 1/2x

    Multiply both sides by 5 to isolate y:

    y = 5(1 - 1/2x)

    Distribute the 5:

    y = 5 - (5/2)x

    Okay, we've isolated y! We now have an expression for y in terms of x.

  2. Substitute into the Other Equation: Now, we take this expression for y and substitute it into the other equation (the one we didn't use in the previous step). This is crucial – if we substitute back into the same equation, we'll just end up with a tautology (something that's always true) and won't solve anything.

    So, we substitute y = 5 - (5/2)x into the second equation:

    1/3x - 1/5y = 4

    Replacing y, we get:

    1/3x - 1/5(5 - (5/2)x) = 4

  3. Simplify and Solve for x: Now we have an equation with only x, so we can solve for it. First, distribute the -1/5:

    1/3x - 1 + 1/2x = 4

    Combine the x terms (we already found a common denominator of 6 when using elimination):

    (2/6)x + (3/6)x - 1 = 4

    (5/6)x - 1 = 4

    Add 1 to both sides:

    (5/6)x = 5

    Multiply both sides by 6/5:

    x = 6

    Woohoo! We got the same value for x as we did with elimination, which is a great sign.

  4. Substitute Back to Find y: Now that we know x = 6, we can substitute it back into the expression we found for y:

    y = 5 - (5/2)x

    y = 5 - (5/2)(6)

    y = 5 - 15

    y = -10

    Awesome! We found the same value for y as before.

The Solution (Again!)

Using the substitution method, we've confirmed that the solution to the system of equations is x = 6 and y = -10, or (6, -10).

Why Substitution Works

Substitution is a powerful method because it allows us to transform a system of two equations into a single equation in one variable. This makes the equation easier to solve. It's like breaking down a complex problem into smaller, more manageable steps.

Choosing the Right Method: Elimination vs. Substitution

So, we've conquered the system of equations using both elimination and substitution. But how do you know which method to use when faced with a new system? Here are a few guidelines:

  • Elimination: This method shines when the coefficients of one of the variables are the same or opposites (or can be easily made that way by multiplying one or both equations by a constant). It's also a good choice when the equations are in standard form (Ax + By = C).
  • Substitution: This method is ideal when one of the equations is already solved for one variable or can be easily solved for one variable. It's also useful when dealing with more complex equations where isolating a variable might be simpler than manipulating coefficients.

In our example, elimination was perhaps slightly faster because the y coefficients were already opposites. However, substitution also worked perfectly well. The key is to practice both methods and develop a feel for which one is best suited for a particular problem.

Let's Recap: Key Takeaways

  • Elimination involves adding or subtracting equations to eliminate one variable.
  • Substitution involves isolating one variable and substituting its expression into the other equation.
  • Both methods will lead to the same solution if applied correctly.
  • Choosing the right method can save you time and effort.

Practice Makes Perfect

The best way to master elimination and substitution is to practice, practice, practice! Try solving different systems of equations using both methods. You'll start to see patterns and develop your problem-solving skills. And remember, math can be fun, especially when you're conquering tough problems!

So, there you have it, folks! We've successfully solved the system of equations 1/2x + 1/5y = 1 and 1/3x - 1/5y = 4 using both elimination and substitution. Keep practicing, and you'll become a system-solving superstar in no time!