Solving Systems Of Equations A Step By Step Guide
Solve the following system of equations:
-3y + 4x = 13
-5y - 6x = -67
```</em></q></p>
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<p>In the realm of mathematics, systems of equations play a crucial role in modeling and solving real-world problems. These systems involve two or more equations with multiple variables, and the goal is to find values for these variables that satisfy all equations simultaneously. In this comprehensive guide, we will delve into the process of solving a system of equations, focusing on a specific example to illustrate the techniques involved. Specifically, we will tackle the following system:</p>
<pre><code class="hljs">-3y + 4x = 13
-5y - 6x = -67
</code></pre>
<p><strong>Solving systems of equations</strong> is a fundamental skill in algebra and has wide-ranging applications in various fields, including engineering, economics, and computer science. Mastering the techniques for solving these systems empowers you to tackle complex problems and gain a deeper understanding of mathematical relationships. Before diving into the solution, let's discuss the significance of solving systems of equations and the different methods available.</p>
<p><strong>The Importance of Solving Systems of Equations</strong></p>
<p>Systems of equations arise naturally in numerous situations where multiple variables are interconnected. For instance, consider a scenario where you need to determine the optimal mix of ingredients for a recipe, given constraints on the total cost and nutritional content. This problem can be formulated as a system of equations, where the variables represent the quantities of each ingredient, and the equations represent the cost and nutritional constraints. Similarly, in physics, systems of equations are used to describe the motion of objects under the influence of multiple forces. In economics, they can model the equilibrium between supply and demand in a market. The ability to solve systems of equations is therefore essential for addressing a wide array of practical problems.</p>
<p><strong>Methods for Solving Systems of Equations</strong></p>
<p>There are several methods available for solving systems of equations, each with its strengths and weaknesses. The most common methods include:</p>
<ol>
<li><strong>Substitution:</strong> This method involves solving one equation for one variable in terms of the other variables and then substituting this expression into the other equations. This reduces the system to a smaller set of equations with fewer variables, which can then be solved using the same technique.</li>
<li><strong>Elimination:</strong> This method involves manipulating the equations in the system to eliminate one or more variables. This is typically achieved by multiplying the equations by suitable constants and then adding or subtracting them. The goal is to obtain a new equation with fewer variables, which can then be solved. The solution is then substituted back into the original equations to solve for the remaining variables.</li>
<li><strong>Graphical Method:</strong> For systems with two variables, the equations can be plotted as lines on a graph. The points of intersection of these lines represent the solutions to the system. This method provides a visual representation of the solutions but may not be accurate for systems with non-linear equations or for finding exact solutions.</li>
<li><strong>Matrix Methods:</strong> For larger systems of linear equations, matrix methods such as Gaussian elimination and matrix inversion provide a systematic and efficient way to find solutions. These methods are particularly useful when dealing with systems with many variables and equations.</li>
</ol>
<p>The choice of method depends on the specific system of equations and the desired level of accuracy. For simple systems with two or three variables, substitution or elimination may be the most straightforward approach. For larger systems, matrix methods may be more efficient.</p>
<p>In the following sections, we will focus on using the <strong>elimination method</strong> to solve the system of equations presented earlier. This method is particularly well-suited for this system, as it allows us to eliminate one of the variables relatively easily.</p>
<p>Now, let's apply the <strong>elimination method</strong> to solve the given system of equations:</p>
<pre><code class="hljs">-3y + 4x = 13 (Equation 1)
-5y - 6x = -67 (Equation 2)
</code></pre>
<p><strong>Step 1: Multiplying Equations to Match Coefficients</strong></p>
<p>The first step in the elimination method is to manipulate the equations so that the coefficients of one of the variables are either equal or additive inverses (i.e., they add up to zero). This allows us to eliminate that variable when we add or subtract the equations. In this case, we can choose to eliminate either <code>x</code> or <code>y</code>. Let's choose to eliminate <code>x</code>. To do this, we need to find a common multiple of the coefficients of <code>x</code> in both equations, which are 4 and -6. The least common multiple of 4 and 6 is 12. Therefore, we will multiply Equation 1 by 3 and Equation 2 by 2 to make the coefficients of <code>x</code> equal to 12 and -12, respectively.</p>
<p>Multiplying Equation 1 by 3, we get:</p>
<pre><code class="hljs">3 * (-3y + 4x) = 3 * 13
-9y + 12x = 39 (Equation 3)
</code></pre>
<p>Multiplying Equation 2 by 2, we get:</p>
<pre><code class="hljs">2 * (-5y - 6x) = 2 * (-67)
-10y - 12x = -134 (Equation 4)
</code></pre>
<p>Now, we have two new equations:</p>
<pre><code class="hljs">-9y + 12x = 39 (Equation 3)
-10y - 12x = -134 (Equation 4)
</code></pre>
<p>Notice that the coefficients of <code>x</code> in these equations are now 12 and -12, which are additive inverses.</p>
<p><strong>Step 2: Eliminating a Variable</strong></p>
<p>Next, we add Equation 3 and Equation 4 to eliminate <code>x</code>:</p>
<pre><code class="hljs">(-9y + 12x) + (-10y - 12x) = 39 + (-134)
-9y + 12x - 10y - 12x = 39 - 134
-19y = -95
</code></pre>
<p>As you can see, the <code>x</code> terms have canceled out, leaving us with a single equation in terms of <code>y</code>.</p>
<p><strong>Step 3: Solving for the Remaining Variable</strong></p>
<p>Now, we solve the equation <code>-19y = -95</code> for <code>y</code> by dividing both sides by -19:</p>
<pre><code class="hljs">y = -95 / -19
y = 5
</code></pre>
<p>So, we have found that <code>y = 5</code>.</p>
<p><strong>Step 4: Substituting to Find the Other Variable</strong></p>
<p>Now that we have the value of <code>y</code>, we can substitute it back into any of the original equations (Equation 1 or Equation 2) to solve for <code>x</code>. Let's substitute <code>y = 5</code> into Equation 1:</p>
<pre><code class="hljs">-3y + 4x = 13
-3(5) + 4x = 13
-15 + 4x = 13
</code></pre>
<p>Now, we solve for <code>x</code>:</p>
<pre><code class="hljs">4x = 13 + 15
4x = 28
x = 28 / 4
x = 7
</code></pre>
<p>So, we have found that <code>x = 7</code>.</p>
<p><strong>Step 5: Verifying the Solution</strong></p>
<p>Finally, it's essential to verify that our solution <code>(x = 7, y = 5)</code> satisfies both original equations. Let's substitute these values into Equation 1 and Equation 2:</p>
<p>For Equation 1:</p>
<pre><code class="hljs">-3y + 4x = 13
-3(5) + 4(7) = 13
-15 + 28 = 13
13 = 13 (Verified)
</code></pre>
<p>For Equation 2:</p>
<pre><code class="hljs">-5y - 6x = -67
-5(5) - 6(7) = -67
-25 - 42 = -67
-67 = -67 (Verified)
</code></pre>
<p>Since the solution <code>(x = 7, y = 5)</code> satisfies both original equations, we have successfully solved the system.</p>
<p>Therefore, the solution to the system of equations is:</p>
<pre><code class="hljs">x = 7
y = 5
</code></pre>
<p>We can also represent the solution as an ordered pair <code>(7, 5)</code>. This indicates that the point <code>(7, 5)</code> is the intersection of the two lines represented by the equations in the system.</p>
<p>While we used the <strong>elimination method</strong> to solve this particular system, it's worth mentioning other methods that can be used to solve systems of equations. These methods include the <strong>substitution method</strong>, the <strong>graphical method</strong>, and <strong>matrix methods</strong>. Let's briefly discuss these alternative approaches.</p>
<p><strong>1. Substitution Method</strong></p>
<p>The <strong>substitution method</strong> involves solving one equation for one variable in terms of the other variable(s) and then substituting this expression into the other equation(s). This reduces the system to a smaller set of equations with fewer variables. For example, we could solve Equation 1 for <code>x</code>:</p>
<pre><code class="hljs">-3y + 4x = 13
4x = 13 + 3y
x = (13 + 3y) / 4
</code></pre>
<p>Then, we would substitute this expression for <code>x</code> into Equation 2:</p>
<pre><code class="hljs">-5y - 6x = -67
-5y - 6((13 + 3y) / 4) = -67
</code></pre>
<p>Now, we have a single equation in terms of <code>y</code>, which we can solve for <code>y</code>. Once we find the value of <code>y</code>, we can substitute it back into the expression for <code>x</code> to find the value of <code>x</code>.</p>
<p><strong>2. Graphical Method</strong></p>
<p>The <strong>graphical method</strong> is suitable for systems with two variables. It involves plotting the equations as lines on a graph. The points of intersection of these lines represent the solutions to the system. For the given system, we can rewrite the equations in slope-intercept form (<code>y = mx + b</code>):</p>
<pre><code class="hljs">Equation 1: -3y + 4x = 13 => y = (4/3)x - 13/3
Equation 2: -5y - 6x = -67 => y = (-6/5)x + 67/5
</code></pre>
<p>We can then plot these lines on a graph and find their point of intersection. The coordinates of this point will give us the values of <code>x</code> and <code>y</code> that satisfy both equations.</p>
<p><strong>3. Matrix Methods</strong></p>
<p>For larger systems of linear equations, <strong>matrix methods</strong> such as Gaussian elimination and matrix inversion provide a systematic and efficient way to find solutions. These methods involve representing the system of equations in matrix form and then performing row operations to solve for the variables. Matrix methods are particularly useful when dealing with systems with many variables and equations.</p>
<p>In this comprehensive guide, we have explored the process of solving a system of equations using the <strong>elimination method</strong>. We tackled a specific example:</p>
<pre><code class="hljs">-3y + 4x = 13
-5y - 6x = -67
</code></pre>
<p>and found the solution to be <code>x = 7</code> and <code>y = 5</code>. We also discussed the importance of solving systems of equations and various methods available, including the substitution method, graphical method, and matrix methods. The <strong>ability to solve systems of equations</strong> is a fundamental skill in mathematics with wide-ranging applications in various fields. By mastering these techniques, you can tackle complex problems and gain a deeper understanding of mathematical relationships.</p>
<p>Remember to always verify your solution by substituting the values back into the original equations to ensure accuracy. With practice and understanding of the different methods, you can confidently solve a wide range of systems of equations.</p>
<p>I hope this guide has been helpful in your understanding of solving systems of equations. Keep practicing and exploring different methods to enhance your skills in this important area of mathematics.</p>