Solving 2x-5y=-14 And 3x+2y=17 With Substitution, Equalization, And Elimination

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Hey guys! Today, we're diving into the exciting world of solving systems of equations. We're going to tackle the following set of equations using three awesome methods: substitution, equalization, and elimination. Buckle up, because this is going to be a mathematical adventure!

The equations we'll be working with are:

  1. 2x - 5y = -14
  2. 3x + 2y = 17

Let's get started!

Method 1: Substitution – The Art of Swapping Variables

Substitution method is one of the most effective ways to solve a system of equations. This method involves solving one equation for one variable and then substituting that expression into the other equation. This eliminates one variable, allowing us to solve for the remaining one. Think of it like a mathematical swap – we're replacing one variable with an equivalent expression. In the substitution method, our initial goal is to isolate one variable in one of the equations. Looking at our equations, let's choose the first equation (2x - 5y = -14) and solve for x. This looks like a good starting point because the coefficient of x is a relatively small number. To isolate x, we first add 5y to both sides of the equation, which gives us 2x = 5y - 14. Then, we divide both sides by 2 to get x = (5y - 14) / 2. Now we have an expression for x in terms of y. The next step is crucial: we substitute this expression for x into the other equation, which is 3x + 2y = 17. This is where the magic of substitution happens. Replacing x with (5y - 14) / 2, we get 3((5y - 14) / 2) + 2y = 17. Notice that this equation now only contains one variable, y. This is exactly what we wanted! To solve for y, we first simplify the equation. Multiply both sides by 2 to get rid of the fraction: 3(5y - 14) + 4y = 34. Expanding the brackets, we have 15y - 42 + 4y = 34. Combining like terms, we get 19y - 42 = 34. Adding 42 to both sides gives us 19y = 76. Finally, dividing both sides by 19, we find that y = 4. Fantastic! We've found the value of y. Now that we know y = 4, we can substitute this value back into either of our original equations or the expression we found for x earlier (x = (5y - 14) / 2). Let's use the expression for x, as it's already set up to solve for x. Substituting y = 4 into x = (5y - 14) / 2, we get x = (5(4) - 14) / 2. Simplifying, we have x = (20 - 14) / 2, which gives us x = 6 / 2, so x = 3. Therefore, the solution to our system of equations using the substitution method is x = 3 and y = 4. We've successfully swapped our way to the answer! This method is particularly powerful when one of the equations can be easily solved for one variable in terms of the other. Remember, the key is to be meticulous with your algebra and keep track of your substitutions.

  1. Solve one equation for one variable. Let's solve the first equation for x:
    • 2x - 5y = -14
    • 2x = 5y - 14
    • x = (5y - 14) / 2
  2. Substitute this expression into the other equation:
    • 3x + 2y = 17
    • 3((5y - 14) / 2) + 2y = 17
  3. Solve for y:
    • (15y - 42) / 2 + 2y = 17
    • 15y - 42 + 4y = 34
    • 19y = 76
    • y = 4
  4. Substitute y back into the expression for x:
    • x = (5(4) - 14) / 2
    • x = (20 - 14) / 2
    • x = 3

So, using the substitution method, we find that x = 3 and y = 4.

Method 2: Equalization – Balancing the Scales of Equations

Equalization method provides another fantastic approach to solving systems of equations. The core idea behind this method is to express the same variable from both equations and then set those expressions equal to each other. This creates a new equation with only one variable, which we can then solve. It's like balancing the scales – we're finding the point where the two expressions for the same variable are equal. To apply the equalization method, the first step is to choose a variable to isolate in both equations. In our system, let's choose to isolate x. It seems like a manageable choice, and we already have x isolated in the first equation from our work with the substitution method. From the first equation, 2x - 5y = -14, we already know that x = (5y - 14) / 2. Now, we need to isolate x in the second equation, 3x + 2y = 17. To do this, we first subtract 2y from both sides, which gives us 3x = 17 - 2y. Then, we divide both sides by 3 to get x = (17 - 2y) / 3. Now we have two expressions for x: x = (5y - 14) / 2 and x = (17 - 2y) / 3. This is where the equalization magic happens! Since both expressions are equal to x, we can set them equal to each other: (5y - 14) / 2 = (17 - 2y) / 3. We've created a new equation with only one variable, y. To solve for y, we first eliminate the fractions by multiplying both sides of the equation by the least common multiple (LCM) of 2 and 3, which is 6. This gives us 3(5y - 14) = 2(17 - 2y). Expanding the brackets, we have 15y - 42 = 34 - 4y. Now, we want to get all the y terms on one side and the constants on the other. Adding 4y to both sides gives us 19y - 42 = 34. Adding 42 to both sides gives us 19y = 76. Finally, dividing both sides by 19, we find that y = 4. Awesome! We've found the value of y using equalization. Now that we know y = 4, we can substitute this value back into either of our expressions for x. Let's use x = (5y - 14) / 2. Substituting y = 4, we get x = (5(4) - 14) / 2. Simplifying, we have x = (20 - 14) / 2, which gives us x = 6 / 2, so x = 3. Therefore, the solution to our system of equations using the equalization method is x = 3 and y = 4. We've successfully balanced our way to the solution! This method is especially handy when both equations can be easily solved for the same variable. The key is to be careful with your algebraic manipulations and make sure you're setting the correct expressions equal to each other. Keep practicing, and you'll become an equalization master!

  1. Solve both equations for the same variable. We already solved the first equation for x:
    • x = (5y - 14) / 2
  2. Solve the second equation for x:
    • 3x + 2y = 17
    • 3x = 17 - 2y
    • x = (17 - 2y) / 3
  3. Set the expressions for x equal to each other:
    • (5y - 14) / 2 = (17 - 2y) / 3
  4. Solve for y:
    • 3(5y - 14) = 2(17 - 2y)
    • 15y - 42 = 34 - 4y
    • 19y = 76
    • y = 4
  5. Substitute y back into either expression for x:
    • x = (5(4) - 14) / 2
    • x = 3

So, with the equalization method, we also get x = 3 and y = 4.

Method 3: Elimination – The Art of Vanishing Variables

Elimination method, also known as the addition or subtraction method, is a powerful technique for solving systems of equations. The basic idea behind the elimination method is to manipulate the equations so that the coefficients of one of the variables are opposites (i.e., one is the negative of the other). Then, when we add the equations together, that variable will be eliminated, leaving us with an equation in just one variable. It’s like making a variable vanish into thin air! The first step in the elimination method is to choose which variable we want to eliminate. Looking at our equations, 2x - 5y = -14 and 3x + 2y = 17, it might be easiest to eliminate y because the coefficients have opposite signs already ( -5 and +2 ). To eliminate y, we need to make the coefficients opposites. The least common multiple (LCM) of 5 and 2 is 10, so we want to make the coefficients of y -10 and +10. To do this, we multiply the first equation by 2 and the second equation by 5. Multiplying the first equation (2x - 5y = -14) by 2, we get 4x - 10y = -28. Multiplying the second equation (3x + 2y = 17) by 5, we get 15x + 10y = 85. Notice that the coefficients of y are now -10 and +10, which are opposites! Now comes the magic step: we add the two new equations together. Adding 4x - 10y = -28 and 15x + 10y = 85, we get (4x + 15x) + (-10y + 10y) = (-28 + 85). This simplifies to 19x = 57. Notice that the y terms have canceled out, leaving us with an equation in just x. To solve for x, we divide both sides by 19, which gives us x = 3. Fantastic! We've found the value of x using elimination. Now that we know x = 3, we can substitute this value back into either of our original equations to solve for y. Let's use the first equation, 2x - 5y = -14. Substituting x = 3, we get 2(3) - 5y = -14. Simplifying, we have 6 - 5y = -14. Subtracting 6 from both sides gives us -5y = -20. Finally, dividing both sides by -5, we find that y = 4. Therefore, the solution to our system of equations using the elimination method is x = 3 and y = 4. We've successfully made a variable vanish and solved for the others! The elimination method is particularly useful when the coefficients of one of the variables are easy to make opposites. The key is to be careful with your multiplication and addition, ensuring that you're accurately eliminating the chosen variable. With practice, you'll be eliminating variables like a pro!

  1. Multiply the equations by constants so that the coefficients of one variable are opposites:
    • Multiply the first equation by 2: 4x - 10y = -28
    • Multiply the second equation by 5: 15x + 10y = 85
  2. Add the equations together to eliminate y:
    • (4x - 10y) + (15x + 10y) = -28 + 85
    • 19x = 57
  3. Solve for x:
    • x = 3
  4. Substitute x back into one of the original equations:
    • 2(3) - 5y = -14
    • 6 - 5y = -14
    • -5y = -20
    • y = 4

And guess what? The elimination method also gives us x = 3 and y = 4!

Conclusion – The Power of Multiple Methods

So, guys, we've successfully solved the system of equations using three different methods: substitution, equalization, and elimination. And the best part? We got the same answer (x = 3, y = 4) with each method! This shows the beauty and consistency of mathematics. Each method offers a unique approach, and choosing the right one often depends on the specific equations you're dealing with. Some systems are easier to solve with substitution, while others lend themselves better to equalization or elimination. The more you practice, the better you'll become at recognizing which method is the most efficient for a given problem. Keep up the awesome work, and you'll conquer any system of equations that comes your way! Remember, math is not just about finding the answer; it's about the journey of problem-solving and the thrill of discovery. So, embrace the challenge, explore different methods, and enjoy the process.

I hope you found this explanation helpful and insightful. Keep practicing and exploring the world of mathematics. You've got this!