Solving Systems Of Equations: Y = -5x + 30 And X = 10

When faced with a system of equations, our goal is to find the specific point, or points, where all the equations in the system hold true simultaneously. Think of it like finding the exact spot on a map where two roads intersect; that intersection is the solution. In this particular case, we have a system with two equations: y = -5x + 30 and x = 10. The question asks for the solution to this system of equations, and we are presented with a few potential answers. To find the solution, we need to determine the values of 'x' and 'y' that satisfy both equations. This process is fundamental in algebra and has numerous applications in fields ranging from economics to physics.

Let's dive into the first equation: y = -5x + 30. This is a linear equation, and it represents a straight line on a graph. The 'y' represents the dependent variable, 'x' represents the independent variable, '-5' is the slope of the line (telling us how steep the line is and its direction), and '+30' is the y-intercept (where the line crosses the y-axis). The second equation, x = 10, is even simpler. It's a vertical line that crosses the x-axis at the value 10. This equation directly tells us the value of 'x' at the point of intersection. It's a crucial piece of information that simplifies the problem significantly. Our task is to take this known value of 'x' and use it to find the corresponding value of 'y' that satisfies both equations. This is where the power of substitution comes into play, a technique that allows us to solve for variables by replacing them with their known equivalents.

To solve this system, the most straightforward method is substitution. We already know from the second equation that x = 10. This means that at the point where our two lines intersect, the x-coordinate must be 10. Now, we can substitute this value of 'x' into the first equation. So, wherever we see an 'x' in the equation y = -5x + 30, we will replace it with the number 10. This gives us a new equation: y = -5(10) + 30. This single equation now only contains the variable 'y', which we can solve for. This is the beauty of substitution – it reduces the complexity of the system, often down to a single variable that can be easily isolated. The process of substitution is a cornerstone of algebraic problem-solving, making complex systems manageable by breaking them down into simpler, solvable parts.

Performing the calculation for 'y' involves a bit of arithmetic. First, we multiply -5 by 10. The product of a negative number and a positive number is negative, so -5 * 10 equals -50. Our equation now becomes y = -50 + 30. Next, we add -50 and 30. When adding numbers with different signs, we find the difference between their absolute values and take the sign of the number with the larger absolute value. The absolute value of -50 is 50, and the absolute value of 30 is 30. The difference between 50 and 30 is 20. Since -50 has a larger absolute value than 30, our result will be negative. Therefore, y = -20. Now we have found both parts of our solution: x = 10 and y = -20. This pair of values represents the coordinates of the point where the lines y = -5x + 30 and x = 10 intersect. It's essential to present the solution as an ordered pair (x, y), which in this case is (10, -20).

Let's verify our solution by plugging these values back into both original equations to ensure they hold true. For the second equation, x = 10, our solution has x = 10, so this equation is satisfied. For the first equation, y = -5x + 30, we substitute x = 10 and y = -20: -20 = -5(10) + 30. We already calculated -5(10) + 30, which simplifies to -50 + 30, resulting in -20. So, -20 = -20. Both equations are satisfied with our solution (10, -20). This verification step is crucial in confirming the accuracy of our work and building confidence in our answer. It's a good habit to always check your answers, especially in mathematics, to catch any potential errors in calculation or logic. This systematic approach to solving and verifying ensures that we arrive at the correct solution with confidence.

Now, let's look at the provided options to see which one matches our calculated solution. The options are: A. (-20,10) B. (10,-20) C. (10,4) D. (4,10)

Comparing our solution, which is (10, -20), with the given options, we can see that option B exactly matches our result. Therefore, the solution to the system of equations is (10, -20). This confirms that our algebraic manipulation and calculations were correct. It's always a good idea to understand why the other options are incorrect as well. For instance, option A (-20,10) would mean x = -20 and y = 10. If we plug x = -20 into the first equation, y = -5(-20) + 30 = 100 + 30 = 130, which is not 10. Option C (10,4) means x = 10 and y = 4. Plugging x=10 into the first equation gives y = -5(10) + 30 = -20, not 4. Option D (4,10) means x = 4 and y = 10. Plugging x=4 into the first equation gives y = -5(4) + 30 = -20 + 30 = 10. This satisfies the first equation, but x=4 does not satisfy the second equation (x=10). This step-by-step elimination of incorrect options further solidifies our understanding and the correctness of our chosen answer. It highlights that only one specific pair of (x, y) coordinates can satisfy both equations simultaneously.

Understanding systems of equations is a fundamental skill in mathematics that opens doors to solving more complex problems. The principles of substitution and elimination, which we used here implicitly through substitution, are powerful tools. For instance, if we had two linear equations without one variable already isolated, we might use the elimination method. This involves manipulating the equations (multiplying them by constants) so that when we add or subtract them, one of the variables cancels out, allowing us to solve for the remaining variable. Graphically, the solution to a system of two linear equations represents the point of intersection of the two lines. If the lines are parallel, there is no solution. If the lines are identical, there are infinitely many solutions. In our case, we have a clear intersection point because the lines are neither parallel nor identical. The equation x=10 represents a vertical line, and y=-5x+30 represents a line with a negative slope, guaranteeing a single intersection point. The visual representation on a graph would show these two lines crossing precisely at the coordinates (10, -20), reinforcing the algebraic solution.

In conclusion, solving systems of equations involves finding the common point(s) that satisfy all equations within the system. For the given system, y = -5x + 30 and x = 10, we successfully used the substitution method. By substituting the known value of x = 10 into the first equation, we calculated y = -20. This yielded the solution (10, -20), which we then verified by plugging it back into both original equations. This rigorous process ensures accuracy and understanding. The ability to solve systems of equations is a vital mathematical competency with broad applications in diverse fields. For further exploration into systems of equations and their various solution methods, you can refer to resources like Khan Academy's comprehensive section on systems of equations, which offers detailed explanations and practice problems.