Solve Systems Of Equations: Same Solution, New Equation!

Hey guys! Ever get stuck on a math problem and wonder if there's a shortcut or a different way to look at it? Well, today we're diving deep into the world of systems of equations, specifically focusing on how to keep the same solution even when you swap out one of the original equations. Think of it like this: you have two pieces of a puzzle that fit together perfectly to give you a unique answer. We're going to see which new puzzle piece can be swapped in without messing up that perfect fit. This is super cool because it shows you the flexibility and underlying logic in algebra.

We're given a system of equations:

$$2x - y = -4$$

$$3x + 5y = 59$$

Our mission, should we choose to accept it, is to find out which of the following equations can replace the second equation ($3x + 5y = 59$) while ensuring the system still has the exact same solution (x, y) as the original system. This means that the point where the two lines represented by the equations intersect remains unchanged. It's all about maintaining that unique intersection point, that one special pair of (x, y) values that makes both original equations true.

Let's break down what it means for a system of equations to have the same solution. When we talk about a system of two linear equations in two variables, like the ones we have here, we're essentially looking at two lines on a graph. The solution to the system is the point where these two lines intersect. If we replace one of the equations with a different one, and the system still has the same solution, it implies that the new equation, along with the unchanged first equation ($2x - y = -4$), also intersect at that exact same point. This can only happen if the new equation is somehow algebraically related to the original second equation in a way that doesn't change the fundamental relationship between x and y that is necessary to satisfy the first equation.

So, how do we figure this out? We've got four options, and we need to test each one. The key principle here is that if an equation can replace another in a system and yield the same solution, it must be linearly dependent on the original equations, or a direct consequence of them. More simply, it means that the new equation must be satisfied by the same (x, y) pair that solves the original system. The easiest way to approach this is often to first find the solution to the original system, and then plug that solution into each of the new equations to see which one holds true. Alternatively, we can look at the options and see if they can be derived from the original equations through valid algebraic operations. For instance, multiplying an equation by a non-zero constant doesn't change its solution set. Adding or subtracting multiples of one equation from another is also a valid operation that maintains the solution set. Let's roll up our sleeves and get to work!

First things first, let's solve the original system to find that magical (x, y) solution. We can use either substitution or elimination. Let's go with elimination. Multiply the first equation ($2x - y = -4$) by 5 to make the y coefficients opposites:

$$5(2x - y) = 5(-4)$$

$$10x - 5y = -20$$

Now, add this modified first equation to the original second equation:

$$(10x - 5y) + (3x + 5y) = -20 + 59$$

$$13x = 39$$

Solving for x:

$$x = \frac{39}{13}$$

$$x = 3$$

Now that we have x, substitute it back into the first original equation ($2x - y = -4$) to find y:

$$2(3) - y = -4$$

$$6 - y = -4$$

Subtract 6 from both sides:

$$-y = -4 - 6$$

$$-y = -10$$

Multiply by -1:

$$y = 10$$

So, the unique solution to the original system is (x, y) = (3, 10). Now, we need to see which of the given options is satisfied by x=3 and y=10. Let's test each option:

Option A: $2x - y = -4$

This is the first equation in our original system. If we replace the second equation with the first one, the system becomes:

$$2x - y = -4$$

$$2x - y = -4$$

This system has infinitely many solutions because the equations are identical. Our goal is to have the same unique solution as the original system, not infinitely many. So, A is not the correct answer because it doesn't maintain the specific unique solution we found, nor does it create a system equivalent to the original two distinct lines. A system of identical equations represents the same line, meaning every point on that line is a solution. That's not what we're after here. We need a system that still points to only (3, 10) as the answer. So, we can immediately rule this out because it fundamentally changes the nature of the solution set from a single point to an entire line.

Option B: $13x = 39$

Let's plug in our solution (3, 10):

$$13(3) = 39$$

$$39 = 39$$

This equation is true for our solution! But does it replace the second equation and maintain the same solution in the context of the system? Remember, for a replacement equation to work, it must, when paired with the other original equation ($2x - y = -4$), produce the same unique solution. If we create a system with $2x - y = -4$ and $13x = 39$, let's see what happens.

From $13x = 39$, we get $x = 3$. Substitute this into $2x - y = -4$:

$$2(3) - y = -4$$

$$6 - y = -4$$

$$y = 10$$

This gives us the solution (3, 10), which is indeed the same unique solution. Now, how did we get $13x = 39$? We derived it by adding a multiple of the first equation to the second equation: $5 imes (2x - y = -4) + (3x + 5y = 59)$. This operation creates a new equation that is a linear combination of the original two. Any equation that can be formed by such a linear combination will be satisfied by the solution of the original system. Crucially, if this new equation ($13x=39$) combined with one of the original equations ($2x-y=-4$) yields the same solution, then it's a valid replacement. Since we showed that pairing $13x=39$ with $2x-y=-4$ gives us (3,10), option B is a strong contender. This demonstrates that $13x = 39$ is a direct consequence of the original system's equations, and thus, it carries the same solution information when combined with one of the original equations.

Option C: $10x - 5y = -20$

Let's plug in our solution (3, 10):

$$10(3) - 5(10) = -20$$

$$30 - 50 = -20$$

$$-20 = -20$$

This equation is also true for our solution! Now, let's look closely at this equation. Notice anything familiar? It's exactly 5 times the first original equation: $5 imes (2x - y = -4) ightarrow 10x - 5y = -20$. If we replace the second equation ($3x + 5y = 59$) with this equation, our new system becomes:

$$2x - y = -4$$

$$10x - 5y = -20$$

This system consists of two equations that are essentially representing the same line, but scaled. The second equation is just the first equation multiplied by 5. Just like in Option A, this results in a system with infinitely many solutions, because both equations describe the exact same relationship between x and y. We need a system that yields the unique solution (3, 10). So, even though (3, 10) satisfies $10x - 5y = -20$, replacing the second equation with it leads to a system with infinite solutions, not the original unique solution. Therefore, C is not the answer.

Option D: $7x = 39$

Let's plug in our solution (3, 10):

$$7(3) = 39$$

$$21 = 39$$

This statement is false. Therefore, the solution (3, 10) does not satisfy this equation, and it cannot be a replacement that maintains the same solution.

Putting it all together:

We found that the original system has the unique solution (3, 10). We then tested each option:

• Option A ($2x - y = -4$) resulted in a system with infinite solutions.
• Option B ($13x = 39$) resulted in a system that, when paired with $2x - y = -4$, produced the unique solution (3, 10).
• Option C ($10x - 5y = -20$) resulted in a system with infinite solutions.
• Option D ($7x = 39$) was not satisfied by the solution (3, 10).

The key insight here is that an equation can replace another in a system if it is a linear combination of the original equations, and when paired with the unchanged original equation, it yields the same unique solution. Option B, $13x = 39$, was derived directly from the original system (it was the result of the elimination step $5 imes ( ext{eq1}) + ( ext{eq2})$), and when paired with the first equation ($2x - y = -4$), it correctly gave us our unique solution (3, 10). This means that $13x = 39$ contains the necessary information to, when combined with $2x-y=-4$, isolate the same unique solution.

Think about what these options represent graphically. The original system represents two distinct lines intersecting at (3, 10). Option A and Option C represent equations that are multiples of the first equation, meaning they represent the same line as the first equation. If you have a system with two identical lines, there are infinite intersection points. Option D represents a line that does not pass through (3, 10). Option B, $13x = 39$, when paired with $2x - y = -4$, also results in lines that intersect at (3, 10). In fact, the equation $13x=39$ itself represents a vertical line at $x=3$. When this vertical line intersects with the line $2x-y=-4$, the intersection point is indeed (3, 10). This confirms that option B is the correct replacement because it forms a new system that is consistent with and yields the same unique solution as the original system.

So, the equation that can replace $3x + 5y = 59$ and still produce the same solution is B. $13x = 39$. This is because this equation is a valid algebraic consequence of the original system, and when paired with the first equation, it leads to the unique solution (3, 10). Pretty neat, huh? It shows how different equations can encode the same mathematical truth about a solution set!

Keep practicing, guys, and you'll become algebra wizards in no time! Understanding these principles makes solving systems of equations much more intuitive and less about rote memorization. It's all about the relationships between the equations and the unique point they define. Peace out!