Solving Systems Of Equations: Find The Solution!

Nov 3, 2025 by Admin 49 views

Hey guys! Today, we're diving into the fascinating world of systems of equations. You know, those sets of equations with multiple variables that look a little intimidating at first glance? But don't worry, we're going to break it down and make it super easy to understand. We'll tackle a specific example to show you exactly how to find the solution. So, buckle up and let's get started!

The Challenge: A System of Equations

Okay, so here's the system of equations we're going to solve:

$\begin{array}{l} 2 x-3 y+z=0 \\ 3 x+2 y=35 \\ 4 y-2 z=14 \end{array}$

Our mission, should we choose to accept it (and we do!), is to find the values of x, y, and z that satisfy all three equations simultaneously. That means when we plug those values into each equation, the equation holds true. Think of it like finding the perfect key that unlocks all three locks at the same time.

We're given a few potential solutions:

A. $(2,3,5)$ B. $(3,2,0)$ C. $(1,16,0)$ D. $(7,7,7)$

So, which one is the correct key? Let's find out!

Method 1: The Substitution Method – A Deep Dive

Let's explore the substitution method. This is a powerful technique where we solve one equation for one variable and then substitute that expression into the other equations. This helps us reduce the number of variables and eventually isolate the values.

Step 1: Simplify and Isolate a Variable

Looking at our system, the third equation, $4y - 2z = 14$ , seems like a good place to start because we can simplify it! Let's divide both sides of the equation by 2:

$2y - z = 7$

Now, let's isolate z by adding it to both sides and subtracting 7 from both sides:

$2y - 7 = z$

Great! We now have an expression for z in terms of y. This is our first key piece of the puzzle.

Step 2: Substitute and Conquer

Now comes the fun part: substitution! We'll take the expression we just found for z ( $z = 2y - 7$ ) and substitute it into the first equation of our system:

$2x - 3y + z = 0$

Replacing z with $(2y - 7)$ , we get:

$2x - 3y + (2y - 7) = 0$

Simplify this equation by combining like terms:

$2x - y - 7 = 0$

Let's rearrange this a bit to isolate the terms with variables on one side:

$2x - y = 7$

Now, we have two equations:

$2x - y = 7$
$3x + 2y = 35$ (the second equation from our original system)

We've effectively reduced our problem to a system of two equations with two variables, x and y. Awesome!

Step 3: Solve the 2x2 System

We can use substitution again, or we can use elimination. Let's use the elimination method this time. To eliminate y, we'll multiply the first equation ( $2x - y = 7$ ) by 2:

$4x - 2y = 14$

Now we have:

$4x - 2y = 14$
$3x + 2y = 35$

Notice that the y terms have opposite signs. Let's add the two equations together. This will eliminate y:

$(4x - 2y) + (3x + 2y) = 14 + 35$

Simplifying, we get:

$7x = 49$

Now, divide both sides by 7 to solve for x:

$x = 7$

Excellent! We've found the value of x.

Step 4: Back-Substitute to Find y and z

Now that we know x = 7, we can substitute this value back into either of the two-variable equations to find y. Let's use the equation $2x - y = 7$ :

$2(7) - y = 7$

$14 - y = 7$

Subtract 14 from both sides:

$-y = -7$

Multiply both sides by -1:

$y = 7$

Fantastic! We've found y = 7.

Finally, let's substitute the value of y into our expression for z ( $z = 2y - 7$ ):

$z = 2(7) - 7$

$z = 14 - 7$

$z = 7$

We did it! We found z = 7.

The Solution!

So, the solution to the system of equations is $(x, y, z) = (7, 7, 7)$ . This corresponds to option D.

Method 2: Testing the Answer Choices – The Quick Route

Sometimes, the fastest way to solve a problem like this is to test the given answer choices. This is especially true when the answer choices are relatively simple numbers. Let's walk through this approach.

The Strategy: Plug and Chug

The idea is simple: we'll take each answer choice (which is a set of values for x, y, and z) and plug those values into the original equations. If the values satisfy all three equations, then that answer choice is the solution. If even one equation is not satisfied, then that answer choice is incorrect.

Testing Option A: (2, 3, 5)

Let's start with option A, $(2, 3, 5)$ . This means x = 2, y = 3, and z = 5. We'll plug these values into each of the three equations:

Equation 1: $2x - 3y + z = 0$ $2(2) - 3(3) + 5 = 4 - 9 + 5 = 0$ (This equation is satisfied!)
Equation 2: $3x + 2y = 35$ $3(2) + 2(3) = 6 + 6 = 12$ (This equation is not satisfied! 12 ≠ 35)

Since option A doesn't satisfy all three equations, we can eliminate it. No need to test the third equation!

Testing Option B: (3, 2, 0)

Next, let's try option B, $(3, 2, 0)$ . So, x = 3, y = 2, and z = 0:

Equation 1: $2x - 3y + z = 0$ $2(3) - 3(2) + 0 = 6 - 6 + 0 = 0$ (Satisfied!)
Equation 2: $3x + 2y = 35$ $3(3) + 2(2) = 9 + 4 = 13$ (Not satisfied! 13 ≠ 35)

Option B is also incorrect.

Testing Option C: (1, 16, 0)

Let's test option C, $(1, 16, 0)$ . Here, x = 1, y = 16, and z = 0:

Equation 1: $2x - 3y + z = 0$ $2(1) - 3(16) + 0 = 2 - 48 + 0 = -46$ (Not satisfied! -46 ≠ 0)

Option C is out!

Testing Option D: (7, 7, 7)

Finally, let's test option D, $(7, 7, 7)$ . This means x = 7, y = 7, and z = 7:

Equation 1: $2x - 3y + z = 0$ $2(7) - 3(7) + 7 = 14 - 21 + 7 = 0$ (Satisfied!)
Equation 2: $3x + 2y = 35$ $3(7) + 2(7) = 21 + 14 = 35$ (Satisfied!)
Equation 3: $4y - 2z = 14$ $4(7) - 2(7) = 28 - 14 = 14$ (Satisfied!)

Option D satisfies all three equations! We found our solution!

The Winner!

Option D, $(7, 7, 7)$ , is the solution to the system of equations.

Key Takeaways for Solving Systems of Equations

Alright, guys, we've conquered this system of equations! Here are some key things to remember when tackling these types of problems:

Substitution Method: This involves solving one equation for one variable and substituting that expression into other equations. It's great for reducing the complexity of the system.
Elimination Method: This involves manipulating equations to eliminate one variable, usually by adding or subtracting multiples of the equations. It's effective when you have coefficients that are easy to work with.
Testing Answer Choices: When you have multiple-choice options, plugging in the values and checking if they satisfy the equations can be a very efficient method, especially when the numbers are relatively simple.
Simplify When Possible: Look for opportunities to simplify equations before you start solving. Dividing both sides by a common factor, for example, can make the numbers smaller and easier to work with.

Practice Makes Perfect

The best way to get comfortable with solving systems of equations is to practice! Try different problems, experiment with different methods, and don't be afraid to make mistakes. Each mistake is a learning opportunity.

So, there you have it! Solving systems of equations doesn't have to be a daunting task. With the right techniques and a little practice, you'll be solving them like a pro in no time. Keep up the great work, and I'll see you in the next math adventure! Remember, math is awesome, and you are awesome for learning it!