Solving Linear Equations A Step By Step Guide

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Hey guys! Today, we're diving into the world of equation solving. It might sound intimidating, but trust me, it's like cracking a code – super satisfying once you get the hang of it. We're going to tackle some common types of equations, break down the steps, and make sure you're equipped to solve them like a pro. So, grab your pencils and let's get started!

(1) Solving $4x = 3x + 5$

When we encounter linear equations like $4x = 3x + 5$, our primary goal is to isolate the variable, which in this case is x. To achieve this, we need to manipulate the equation using algebraic operations, ensuring that we maintain the balance on both sides. The key here is to think of the equation as a balanced scale; whatever you do on one side, you must do on the other to keep it level.

First, let's gather all the x terms on one side of the equation. In this instance, we have 4x on the left and 3x on the right. To consolidate these, we can subtract 3x from both sides. This operation is based on the subtraction property of equality, which states that if you subtract the same value from both sides of an equation, the equality holds true. Subtracting 3x from both sides gives us:

4x−3x=3x+5−3x4x - 3x = 3x + 5 - 3x

This simplifies to:

x=5x = 5

And just like that, we've solved for x! The equation now clearly shows that x is equal to 5. To ensure our solution is correct, it's always a good practice to verify the answer. We do this by substituting the value of x back into the original equation.

Substituting x = 5 into $4x = 3x + 5$, we get:

4(5)=3(5)+54(5) = 3(5) + 5

20=15+520 = 15 + 5

20=2020 = 20

The equation holds true, confirming that our solution x = 5 is indeed correct. This process of substitution is a crucial step in problem-solving, as it helps eliminate potential errors and builds confidence in the result. When teaching this concept, it's essential to emphasize the importance of checking answers to foster accuracy and a deeper understanding of the equation-solving process. The act of verification not only confirms the solution but also reinforces the student's grasp of algebraic manipulations and the properties of equality.

(2) Tackling $7m = 8 + 5m$

Alright, let's move on to our next equation: $7m = 8 + 5m$. Similar to the previous example, our mission is to isolate the variable, which in this case is m. To do this effectively, we'll employ the same principles of algebraic manipulation, focusing on maintaining the equation's balance. The key is to remember that whatever operation we perform on one side of the equation, we must also perform on the other side to preserve the equality.

Our first step should be to gather all the m terms on one side of the equation. Looking at the equation, we have 7m on the left side and 5m on the right side. To consolidate these terms, we can subtract 5m from both sides. This step utilizes the subtraction property of equality, a fundamental concept in algebra. Subtracting 5m from both sides, we get:

7m−5m=8+5m−5m7m - 5m = 8 + 5m - 5m

This simplifies to:

2m=82m = 8

Now, we have 2m on one side and a constant on the other. To isolate m, we need to get rid of the coefficient 2. We can do this by dividing both sides of the equation by 2. This step is based on the division property of equality, which allows us to divide both sides of an equation by the same non-zero number without changing the equality. Dividing both sides by 2, we have:

2m2=82\frac{2m}{2} = \frac{8}{2}

This simplifies to:

m=4m = 4

Fantastic! We've found the value of m. But, just like before, it's crucial to verify our solution to ensure its accuracy. We'll substitute m = 4 back into the original equation: $7m = 8 + 5m$.

Substituting m = 4, we get:

7(4)=8+5(4)7(4) = 8 + 5(4)

28=8+2028 = 8 + 20

28=2828 = 28

The equation holds true, confirming that our solution m = 4 is correct. This verification step is not just a formality; it's an essential part of the problem-solving process. It helps to catch any mistakes made during the algebraic manipulations and reinforces the understanding of how equations work. Encouraging students to always check their solutions is a great way to build their confidence and accuracy in solving equations.

(3) Solving $2a = 9 - a$

Moving on, let's tackle the equation $2a = 9 - a$. As you might have guessed, our main objective remains the same: to isolate the variable, which in this case is a. To achieve this, we'll continue to use algebraic manipulations while ensuring the equation stays balanced. Remember, whatever action we take on one side, we must mirror on the other side to maintain the equality.

Our initial step should be to gather all the a terms on one side of the equation. Looking at the equation, we have 2a on the left side and -a on the right side. To bring these terms together, we can add a to both sides. This step is justified by the addition property of equality, which states that adding the same value to both sides of an equation preserves the equality. Adding a to both sides, we get:

2a+a=9−a+a2a + a = 9 - a + a

This simplifies to:

3a=93a = 9

Now, we have 3a on one side and a constant on the other. To isolate a, we need to eliminate the coefficient 3. We can do this by dividing both sides of the equation by 3. This step is based on the division property of equality, which we discussed earlier. Dividing both sides by 3, we have:

3a3=93\frac{3a}{3} = \frac{9}{3}

This simplifies to:

a=3a = 3

Excellent! We've found the value of a. Now, let's not forget our crucial step of verifying the solution. We'll substitute a = 3 back into the original equation: $2a = 9 - a$.

Substituting a = 3, we get:

2(3)=9−32(3) = 9 - 3

6=66 = 6

The equation holds true, confirming that our solution a = 3 is indeed correct. This consistent practice of verification is what separates good problem-solvers from great ones. It not only ensures accuracy but also deepens the understanding of the underlying algebraic principles. When teaching these concepts, emphasizing the importance of this verification step can significantly enhance students' ability to solve equations confidently and accurately.

(4) Solving $ rac{1}{4}d = 3$

Alright, let's switch gears slightly and look at the equation $ rac{1}{4}d = 3$. In this equation, we're dealing with a fractional coefficient for our variable, d. But don't worry, the same principles of isolating the variable apply here. Our goal remains to get d by itself on one side of the equation.

In this case, d is being multiplied by $\frac{1}{4}$. To undo this multiplication and isolate d, we can multiply both sides of the equation by the reciprocal of $\frac{1}{4}$, which is 4. This operation is based on the multiplication property of equality, which states that multiplying both sides of an equation by the same non-zero number maintains the equality. Multiplying both sides by 4, we get:

4â‹…14d=4â‹…34 \cdot \frac{1}{4}d = 4 \cdot 3

This simplifies to:

d=12d = 12

And there you have it! We've solved for d. But, as always, we need to verify our solution to make sure it's correct. We'll substitute d = 12 back into the original equation: $\frac{1}{4}d = 3$.

Substituting d = 12, we get:

14(12)=3\frac{1}{4}(12) = 3

3=33 = 3

The equation holds true, confirming that our solution d = 12 is correct. This example highlights the importance of understanding how to deal with fractions in equations. By multiplying by the reciprocal, we efficiently isolated the variable. Emphasizing this technique to students can help them tackle similar problems with greater ease and confidence. Remember, the key to mastering equation solving is understanding the properties of equality and applying them consistently.

(5) Solving $5y + 6 = 21$

Now, let's move on to solving the equation $5y + 6 = 21$. This equation introduces a new element: a constant term added to the variable term. But don't fret, we'll tackle this step-by-step using our trusty algebraic principles. Our main goal, as always, is to isolate the variable, which in this case is y.

Our first step should be to isolate the term containing y. To do this, we need to get rid of the constant term, which is +6. We can achieve this by subtracting 6 from both sides of the equation. This is an application of the subtraction property of equality, which we've used before. Subtracting 6 from both sides, we get:

5y+6−6=21−65y + 6 - 6 = 21 - 6

This simplifies to:

5y=155y = 15

Now, we have 5y on one side and a constant on the other. To isolate y, we need to eliminate the coefficient 5. We can do this by dividing both sides of the equation by 5. This step is based on the division property of equality. Dividing both sides by 5, we have:

5y5=155\frac{5y}{5} = \frac{15}{5}

This simplifies to:

y=3y = 3

Great! We've found the value of y. But remember, we always need to verify our solution to ensure its accuracy. We'll substitute y = 3 back into the original equation: $5y + 6 = 21$.

Substituting y = 3, we get:

5(3)+6=215(3) + 6 = 21

15+6=2115 + 6 = 21

21=2121 = 21

The equation holds true, confirming that our solution y = 3 is indeed correct. This example illustrates the importance of dealing with constant terms before addressing coefficients. By systematically applying the properties of equality, we can solve even slightly more complex equations with confidence. Emphasizing this order of operations to students can help them develop a methodical approach to problem-solving.

(6) Solving $2n - 3 = 6$

Let's jump into another equation: $2n - 3 = 6$. This equation is similar to the previous one, with a constant term involved, but this time it's being subtracted from the variable term. Our goal remains consistent: to isolate the variable, which in this case is n. We'll continue to use our algebraic tools and principles to achieve this.

Our first step should be to isolate the term containing n. To do this, we need to get rid of the constant term, which is -3. To counteract this subtraction, we can add 3 to both sides of the equation. This is an application of the addition property of equality. Adding 3 to both sides, we get:

2n−3+3=6+32n - 3 + 3 = 6 + 3

This simplifies to:

2n=92n = 9

Now, we have 2n on one side and a constant on the other. To isolate n, we need to eliminate the coefficient 2. We can do this by dividing both sides of the equation by 2. This step is based on the division property of equality. Dividing both sides by 2, we have:

2n2=92\frac{2n}{2} = \frac{9}{2}

This simplifies to:

n=92n = \frac{9}{2}

Or, if we prefer a decimal representation, n = 4.5. Fantastic! We've found the value of n. Now, let's not forget our crucial step of verifying the solution. We'll substitute n = 9/2 (or n = 4.5) back into the original equation: $2n - 3 = 6$.

Substituting n = 9/2, we get:

2(92)−3=62(\frac{9}{2}) - 3 = 6

9−3=69 - 3 = 6

6=66 = 6

The equation holds true, confirming that our solution n = 9/2 (or n = 4.5) is indeed correct. This example reinforces the importance of handling constant terms and coefficients systematically. By applying the properties of equality in the correct order, we can confidently solve for the variable, even when the solution is a fraction or a decimal. Emphasizing this systematic approach to students can help them build a strong foundation in equation solving.

(7) Solving $4 + 3x = 17$

Let's tackle another equation: $4 + 3x = 17$. This equation is similar to some of our previous examples, with a constant term and a variable term. Our trusty objective remains the same: to isolate the variable, which in this case is x. We'll continue to employ our algebraic techniques to achieve this goal.

Our first step should be to isolate the term containing x. To do this, we need to get rid of the constant term, which is +4. We can achieve this by subtracting 4 from both sides of the equation. This is an application of the subtraction property of equality. Subtracting 4 from both sides, we get:

4+3x−4=17−44 + 3x - 4 = 17 - 4

This simplifies to:

3x=133x = 13

Now, we have 3x on one side and a constant on the other. To isolate x, we need to eliminate the coefficient 3. We can do this by dividing both sides of the equation by 3. This step is based on the division property of equality. Dividing both sides by 3, we have:

3x3=133\frac{3x}{3} = \frac{13}{3}

This simplifies to:

x=133x = \frac{13}{3}

Or, if we prefer a decimal representation, x ≈ 4.33. Awesome! We've found the value of x. Now, let's not forget our crucial step of verifying the solution. We'll substitute x = 13/3 (or x ≈ 4.33) back into the original equation: $4 + 3x = 17$.

Substituting x = 13/3, we get:

4+3(133)=174 + 3(\frac{13}{3}) = 17

4+13=174 + 13 = 17

17=1717 = 17

The equation holds true, confirming that our solution x = 13/3 (or x ≈ 4.33) is indeed correct. This example further emphasizes the importance of a systematic approach to equation solving. By consistently applying the properties of equality, we can confidently handle equations with fractional or decimal solutions. Encouraging students to verify their solutions, as we've done here, is a great way to build their accuracy and understanding.

(8) Solving $15 = 4a + 3$

Let's dive into our final equation: $15 = 4a + 3$. This equation looks slightly different because the constant term is on the left side, but don't let that throw you off! Our ultimate goal remains the same: to isolate the variable, which in this case is a. We'll continue to use our algebraic skills to achieve this.

Our first step should be to isolate the term containing a. To do this, we need to get rid of the constant term on the same side, which is +3. We can achieve this by subtracting 3 from both sides of the equation. This is, once again, an application of the subtraction property of equality. Subtracting 3 from both sides, we get:

15−3=4a+3−315 - 3 = 4a + 3 - 3

This simplifies to:

12=4a12 = 4a

Now, we have a constant on one side and 4a on the other. To isolate a, we need to eliminate the coefficient 4. We can do this by dividing both sides of the equation by 4. This step is based on the division property of equality. Dividing both sides by 4, we have:

124=4a4\frac{12}{4} = \frac{4a}{4}

This simplifies to:

3=a3 = a

Which we can also write as a = 3. Excellent! We've found the value of a. Now, as always, we need to verify our solution to ensure its accuracy. We'll substitute a = 3 back into the original equation: $15 = 4a + 3$.

Substituting a = 3, we get:

15=4(3)+315 = 4(3) + 3

15=12+315 = 12 + 3

15=1515 = 15

The equation holds true, confirming that our solution a = 3 is indeed correct. This example demonstrates that the position of the terms in the equation doesn't change our approach. We consistently apply the properties of equality to isolate the variable. Encouraging students to be flexible and adaptable in their problem-solving strategies is key to their success in algebra.

Conclusion: Mastering Equations

And there you have it, guys! We've solved a variety of equations, from simple ones to those with fractions and constant terms. Remember, the key to solving equations is understanding the properties of equality and applying them systematically. Always isolate the variable, and never forget to verify your solutions. With practice, you'll become equation-solving pros in no time! Keep up the great work!