Solve For X And Y: System Of Equations Explained
Hey guys! Let's dive into solving a system of equations. It's like a puzzle where we need to find the values of our variables, in this case, x and y, that make both equations true. We're given two equations:
- 3x + 2y = 14
- x = 4y - 2
Our mission, should we choose to accept it (and we do!), is to find the x and y values that satisfy both of these equations simultaneously. There are a couple of cool methods we can use to crack this code: substitution and elimination. Let's explore both!
Method 1: Substitution – The Sneaky Switcheroo
The substitution method is like a clever switcheroo. We isolate one variable in one equation and then substitute that expression into the other equation. This leaves us with a single equation with just one variable, which we can easily solve. Then, we plug that value back into one of the original equations to find the other variable. Let’s see this in action!
In our system, the second equation (x = 4y - 2) is already solved for x. How convenient! This makes substitution super easy. We're going to take that expression for x (4y - 2) and substitute it into the first equation wherever we see an x. So, the first equation (3x + 2y = 14) transforms into:
3(4y - 2) + 2y = 14
Now, we have a single equation with just y as the variable. Time to unleash our algebraic skills! First, we distribute the 3:
12y - 6 + 2y = 14
Next, we combine like terms:
14y - 6 = 14
Now, we isolate the y term by adding 6 to both sides:
14y = 20
Finally, we solve for y by dividing both sides by 14:
y = 20/14
We can simplify this fraction by dividing both the numerator and denominator by their greatest common divisor, which is 2:
y = 10/7
Awesome! We've found the value of y. But we're not done yet – we still need to find x. Remember that second equation, x = 4y - 2? Now's its time to shine! We'll plug our value of y (10/7) into this equation to solve for x:
x = 4(10/7) - 2
x = 40/7 - 2
To subtract 2, we need to express it as a fraction with a denominator of 7:
x = 40/7 - 14/7
x = 26/7
Boom! We've found both x and y! Our solution is x = 26/7 and y = 10/7. This means the ordered pair (x, y) that satisfies both equations is (26/7, 10/7).
Method 2: Elimination – The Art of Cancellation
The elimination method is like a strategic cancellation. We manipulate the equations so that when we add them together, one of the variables disappears. This leaves us with a single equation with one variable, just like in substitution. Let’s see how it works with our system:
- 3x + 2y = 14
- x = 4y - 2
First, we need to rearrange the second equation so that the x and y terms are on the same side:
x - 4y = -2
Now, our system looks like this:
- 3x + 2y = 14
- x - 4y = -2
To eliminate a variable, we need the coefficients of either x or y to be opposites. Let's focus on eliminating y. The coefficient of y in the first equation is 2, and in the second equation, it's -4. To make them opposites, we can multiply the first equation by 2. This will give us a coefficient of 4 for y in the first equation and -4 for y in the second equation.
Multiplying the first equation by 2, we get:
6x + 4y = 28
Now, our system is:
- 6x + 4y = 28
- x - 4y = -2
Now comes the fun part – elimination! We add the two equations together. Notice what happens to the y terms:
(6x + 4y) + (x - 4y) = 28 + (-2)
6x + x + 4y - 4y = 26
7x = 26
The y terms have canceled out! We're left with a simple equation in x. We solve for x by dividing both sides by 7:
x = 26/7
We've found x! Now, we need to find y. We can plug this value of x into either of the original equations to solve for y. Let's use the second equation, x = 4y - 2:
26/7 = 4y - 2
Add 2 to both sides:
26/7 + 2 = 4y
To add 2, we need to express it as a fraction with a denominator of 7:
26/7 + 14/7 = 4y
40/7 = 4y
Divide both sides by 4 (which is the same as multiplying by 1/4):
(40/7) * (1/4) = y
10/7 = y
We've found y! Our solution using elimination is x = 26/7 and y = 10/7, which matches the solution we found using substitution. Hooray!
Checking Our Work – The Final Sanity Check
It's always a good idea to check our solution to make sure we haven't made any mistakes. We'll plug our values of x and y (x = 26/7, y = 10/7) back into both original equations to see if they hold true.
Let's start with the first equation:
3x + 2y = 14
3(26/7) + 2(10/7) = 14
78/7 + 20/7 = 14
98/7 = 14
14 = 14
The first equation checks out! Now, let's check the second equation:
x = 4y - 2
26/7 = 4(10/7) - 2
26/7 = 40/7 - 2
26/7 = 40/7 - 14/7
26/7 = 26/7
The second equation checks out too! This confirms that our solution, x = 26/7 and y = 10/7, is indeed correct.
Conclusion – System Solved!
We've successfully navigated the world of systems of equations and found the values of x and y that satisfy both equations. We explored two powerful methods: substitution and elimination. Both methods led us to the same solution, giving us confidence in our answer. Remember, solving systems of equations is a fundamental skill in algebra, and mastering it opens doors to tackling more complex mathematical problems. Keep practicing, and you'll become a system-solving pro in no time!
