1.1: Linear Equations
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A linear equation is an equation where the highest exponent on the given variables is one. A linear equation in one variable is an equation with one variable with exponent one, e.g., \[ax+b=c,\nonumber\] where \(a\) is called the coefficient of \(x\), and \(b\) and \(c\) are constant coefficients.
Solving linear equations is an important and fundamental skill in algebra. In algebra, we are often presented with a problem where the answer is unknown. This is usually represented by a variable. There will be times when there are multiple unknowns and we use algebra techniques to solve for the variable.
Be sure to understand the difference between an expression and an equation.
Recall. We simplify expressions and solve equations. Hence, expressions do not contain an equal sign, \(=\), and we only distribute and combine like terms. Equations contain an equal sign, \(=\), and we solve for the variable in addition to distributing and combining like terms.
The result to an expression is an expression and the result to an equation is a number (with the exception of when the result is infinitely many solutions or no solution).
Verifying Solutions
Is \(x = −5\) a solution to the equation \(4x + 16 = −4\)?
Solution
We substitute \(x = −5\) into the equation and determine if the left side is equal to the right side.
\[\begin{array}{rl}4(-5)+16\stackrel{?}{=}-4&\text{Multiply }4(-5) \\ -20+16\stackrel{?}{=}-4&\text{Add }-20+16 \\ -4=-4&\checkmark\text{ True}\end{array}\nonumber\]
Hence, \(x = −5\) is a solution to the equation \(4x + 16 = −4\).
Is \(x = 3\) a solution to the equation \(4x + 16 = −4\)?
Solution
We substitute \(x = 3\) into the equation and determine if the left side is equal to the right side.
\[\begin{array}{rl}4(3)+16\stackrel{?}{=}-4&\text{Multiply }4(3) \\ 12+16\stackrel{?}{=}-4&\text{Add }12+16 \\ 28\neq -4&X\text{ False}\end{array}\nonumber\]
Hence, \(x = 3\) is not a solution to the equation \(4x + 16 = −4\).
Example \(\PageIndex{2}\) reiterates that the solution to each equation is unique. Once we see that at one solution makes the equation true, then we look no further. The exception is when we have an identity, which we discuss later in this chapter.
One-Step Equations
Depending on the complexity of the problem, this “guess and check” method is not very efficient. Thus, we take a more algebraic approach for solving equations. Here we will focus on equations that only require one step to solve. While these equations often seem very fundamental, it is important to master the pattern for solving these problems so we can solve more complex problems.
Given an equation, \(a = b\), the following is an equivalent statement: \[a+c=b+c\nonumber\] i.e., we can add any number to one side of the equation as long as we add the same number to the other side. Often, we use this property to isolate the variable.
Solve for \(x\): \(x + 7 = −5\)
Solution
\[\begin{array}{rl}x+7=-5&\text{Isolate }y\text{ by adding the opposite of }7 \\ x+7+\color{blue}{(-7)}\color{black}{}=-5+\color{blue}{(-7)}\color{black}{}&\text{Simplify} \\ x=-12&\text{Solution}\end{array}\nonumber\]
Recall, it is encouraged for the student to check the obtained answer by verifying the solution:
\[\begin{array}{rl}\color{blue}{(-12)}\color{black}{}+7\stackrel{?}{=}-5 \\ -5=-5&\checkmark\text{ True}\end{array}\nonumber\]
Solve for \(y\): \(4 + y = 8\)
Solution
\[\begin{array}{rl}4+y=8&\text{Isolate }y\text{ by adding the opposite of }4 \\ 4+y+\color{blue}{(-4)}\color{black}{}=8+\color{blue}{(-4)}\color{black}{}&\text{Simplify} \\ y=4&\text{Solution}\end{array}\nonumber\]
Solve for \(y\): \(7 = y + 9\)
Solution
\[\begin{array}{rl}7=y+9&\text{Isolate }y\text{ by adding the opposite of }9 \\ 7+\color{blue}{(-9)}\color{black}{}=y+9+\color{blue}{(-9)}\color{black}{}&\text{Simplify} \\ -2=y&\text{Rewrite with }y\text{ on the left side} \\ y=-2&\text{Solution}\end{array}\nonumber\]
Solve for \(x\): \(5 = 8 + x\)
Solution
\[\begin{array}{rl}5=8+x&\text{Isolate }x\text{ by adding the opposite of }8 \\ 5+\color{blue}{(-8)}\color{black}{}=8+x+\color{blue}{(-8)}\color{black}{}&\text{Simplify} \\ -3=x&\text{Rewrite with }x\text{ on the left side} \\ x=-3&\text{Solution}\end{array}\nonumber\]
Be sure to check your answer by verifying the solution! It only take a few seconds and will save you time and future common errors in the long run.
Solve for \(y\): \(y − 5 = 4\)
Solution
\[\begin{array}{rl}y-5=4&\text{Isolate }y\text{ by adding the opposite of }-5 \\ y-5+\color{blue}{(5)}\color{black}{}=4+\color{blue}{(5)}\color{black}{}&\text{Simplify} \\ y=9&\text{Solution}\end{array}\nonumber\]
Solve for \(y\): \(−10 = y − 7\)
Solution
\[\begin{array}{rl}-10=y-7&\text{Isolate }y\text{ by adding the opposite of }-7 \\ -10+\color{blue}{(7)}\color{black}{}=y-7+\color{blue}{(7)}\color{black}{}&\text{Simplify} \\ y=-3&\text{Solution}\end{array}\nonumber\]
Given an equation, \(a = b\), the following is an equivalent statement: \[ac=bc,\nonumber\] where \(c\neq 0\), i.e., we can multiply both sides of the equation by a nonzero number. Often, we use this property to isolate the variable when there is a coefficient in front of the variable.
Solve for \(y\): \(4y = 20\)
Solution
\[\begin{array}{rl}4y=20&\text{Isolate }y\text{ by multiplying by the reciprocal of }4 \\ \color{blue}{\frac{1}{4}}\color{black}{}\cdot 4y=20\cdot\color{blue}{\frac{1}{4}}\color{black}{}&\text{Simplify} \\ y=5&\text{Solution}\end{array}\nonumber\]
In Example \(\PageIndex{9}\), we multiplied by the reciprocal. The product of a number and its reciprocal is one, i.e., if \(c\) is a number, then its reciprocal is \(\frac{1}{c}\) because \[c\cdot\frac{1}{c}=1\nonumber\]
Solve for \(y\): \(−5y = 30\)
Solution
\[\begin{array}{rl}-5y=30&\text{Isolate }y\text{ by multiplying by the reciprocal of }-5 \\ \color{blue}{-\frac{1}{5}}\color{black}{}\cdot -5y=30\cdot\color{blue}{-\frac{1}{5}}\color{black}{}&\text{Simplify} \\ y=-6&\text{Solution}\end{array}\nonumber\]
Solve for \(x\): \(−42 = −7x\)
Solution
\[\begin{array}{rl}-42=-7x&\text{Isolate }x\text{ by multiplying by the reciprocal of }-7 \\ \color{blue}{-\frac{1}{7}}\color{black}{}\cdot -42=-7x\cdot\color{blue}{-\frac{1}{7}}\color{black}{}&\text{Simplify} \\ 6=x&\text{Rewrite with }x\text{ on the left side} \\ x=6&\text{Solution}\end{array}\nonumber\]
Solve for \(x\): \(\frac{x}{-7}=-2\)
Solution
\[\begin{array}{rl}\frac{x}{-7}=-2&\text{Isolate }x\text{ by multiplying by the reciprocal of }-\frac{1}{7} \\ \color{blue}{-7}\color{black}{}\cdot\frac{x}{-7}=-2\cdot\color{blue}{-7}\color{black}{}&\text{Simplify} \\ x=14&\text{Solution}\end{array}\nonumber\]
Solve for \(x\): \(\frac{x}{5}=-3\)
Solution
\[\begin{array}{rl}\frac{x}{5}=-3&\text{Isolate }x\text{ by multiplying by the reciprocal of }\frac{1}{5} \\ \color{blue}{5}\color{black}{}\cdot\frac{x}{5}=-3\cdot\color{blue}{5}\color{black}{}&\text{Simplify} \\ x=-15&\text{Solution}\end{array}\nonumber\]
The processes described above is fundamental for solving equations. Once these processes are mastered, we are given problems that involve multiple steps. These problems may seem more complex, but the processes and patterns are the same.
The study of algebra was originally called “Cossic Art” from the Latin, meaning the study of “things,” which we now call variables.
Two-Step Equations
After mastering the techniques for solving one-step equations, we are ready to consider two-step equations. As we solve two-step equations, the important thing to remember is that everything works in reverse. When working with one-step equations, we learned that in order to clear a “plus five” in the equation, we would subtract five (or add its opposite) from both sides. We learned that to clear “divided by seven,” we multiply by seven (or multiply by its reciprocal) on both sides. When solving for our variable, we use order of operations in reverse. This means we will apply the addition property first, then the multiplication property second (then exponents, and, finally, any parenthesis or grouping symbols, but that’s another lesson).
Solve for \(y\): \(4y − 20 = −8\)
Solution
We have one extra term on the same side as the variable term, \(4y\). We will first isolate the variable term, then isolate the variable by applying the properties of equations: \[\begin{array}{rl}4y-20=-8&\text{Add the opposite of }-20\text{ to each side} \\ 4y-20+\color{blue}{20}\color{black}{}=-8+\color{blue}{20}\color{black}{}&\text{Simplify} \\ 4y=12&\text{Multiply by the reciprocal of }4 \\ \color{blue}{\frac{1}{4}}\color{black}{}\cdot 4y=12\cdot\color{blue}{\frac{1}{4}}\color{black}{}&\text{Simplify} \\ y=3&\text{Solution}\end{array}\nonumber\]
Let's verify the solution: \[\begin{array}{rl}4\color{blue}{(3)}\color{black}{}-20\stackrel{?}{=}-8 \\ 12-20\stackrel{?}{=}-8 \\ -8=-8&\checkmark\text{ True}\end{array}\nonumber\]
Thus, \(y = 3\) is the solution to the equation.
The same process is used to solve any two-step equations. Add or subtract first, then multiply or divide. Recall, the method never changes, just problems do.
Solve for \(w\): \(5w + 7 = 7\)
Solution
\[\begin{array}{rl}5w+7=7&\text{Add the opposite of }7\text{ to each side} \\ 5w+7+\color{blue}{(-7)}\color{black}{}=7+\color{blue}{(-7)}\color{black}{}&\text{Simplify} \\ 5w=0&\text{Multiply by the reciprocal of }5 \\ \color{blue}{\frac{1}{5}}\color{black}{}\cdot 5w=0\cdot\color{blue}{\frac{1}{5}}\color{black}{}&\text{Simplify} \\ w=0&\text{Solution}\end{array}\nonumber\]
Notice in Example \(\PageIndex{15}\) the seven canceled out completely. Many students get stuck at this point. Let’s not forget that the product of a number and zero is zero. Hence, zero is the solution to the equation.
A common error students make with two-step equations is with negative signs. Remember, the sign always stays with the coefficient.
Solve for \(t\): \(4-2t=10\)
Solution
\[\begin{array}{rl}4-2t=10&\text{Add the opposite of }4\text{ to each side} \\ 4-2t+\color{blue}{(-4)}\color{black}{}=10+\color{blue}{(-4)}\color{black}{}&\text{Simplify} \\ -2t=6&\text{Multiply by the reciprocal of }-2 \\ \color{blue}{-\frac{1}{2}}\color{black}{}\cdot -2t=6\cdot\color{blue}{-\frac{1}{2}}\color{black}{}&\text{Simplify} \\ t=-3&\text{Solution}\end{array}\nonumber\]
Solve for \(n\): \(8 − n = 2\)
Solution
\[\begin{array}{rl}8-n=2&\text{Add the opposite of }8\text{ to each side} \\ 8-n+\color{blue}{(-8)}\color{black}{}=2+\color{blue}{(-8)}\color{black}{}&\text{Simplify} \\ -n=-6&\text{Rewrite }-n\text{ as }-1n \\ -1n=-6&\text{Multiply by the reciprocal of }-1 \\ \color{blue}{-1}\color{black}{}\cdot -1n=-6\cdot\color{blue}{-1}\color{black}{}&\text{Simplify} \\ n=6&\text{Solution}\end{array}\nonumber\]
Solve for \(y\): \(−3y + 7 = −8\)
Solution
\[\begin{array}{rl}-3y+7=-8&\text{Add the opposite of }7\text{ to each side} \\ -3y+7+\color{blue}{(-7)}\color{black}{}=-8+\color{blue}{(-7)}\color{black}{}&\text{Simplify} \\ -3y=-15&\text{Multiply by the reciprocal of }-3 \\ \color{blue}{-\frac{1}{3}}\color{black}{}\cdot -3y=-15\cdot\color{blue}{-\frac{1}{3}}\color{black}{}&\text{Simplify} \\ y=5&\text{Solution}\end{array}\nonumber\]
Solve for \(x\): \(−3 = \frac{x}{5} − 4\)
Solution
Notice the variable term is on the right side of the equation. However, we still follow the method as usual: \[\begin{array}{rl}-3=\frac{x}{5}-4&\text{Add the opposite of }-4\text{ to each side} \\ -3+\color{blue}{4}\color{black}{}=\frac{x}{5}-4+\color{blue}{4}\color{black}{}&\text{Simplify} \\ 1=\frac{x}{5}&\text{Multiply by the reciprocal of }\frac{1}{5} \\ \color{blue}{5}\color{black}{}\cdot 1=\frac{x}{5}\cdot\color{blue}{5}\color{black}{}&\text{Simplify} \\ 5=x&\text{Rewrite the solution with the variable on the left side} \\ x=5&\text{Solution}\end{array}\nonumber\]
As problems in algebra become more complex, the method remains the same. In fact, as we solve problems like those in the next example, each one of them will have several steps to solve, but the last two steps are a two-step equation. This is a critical reason to master two-step equations.
\[3x^2+4=y+6\qquad \frac{1}{x-8}+\frac{1}{x}=\frac{1}{3}\qquad \sqrt{5y-5}+1=y\qquad\log _5(2y-4)=1\nonumber\]
Persian mathematician Omar Khayyam would solve algebraic problems geometrically by intersecting graphs rather than solving them algebraically.
General Equations
Often as we are solving linear equations, we will need to do some preliminary work so that it is more familiar for us to solve. This section will focus on manipulating an equation in such a way that we can use our method for solving two-step equations to obtain the solution.
One such issue that needs to be addressed is parentheses. Sometimes parenthesis can get in the way of solving an equation. As you might expect, we can rewrite an equation without parenthesis by using the distributive property.
The following is an equivalent statement: \[c(a+b)=ca+cb\nonumber\] where \(c\neq 0\), i.e., we can multiply each term in the parenthesis by a nonzero number.
Solve for \(y\): \(4(2y − 6) = 16\)
Solution
Notice the first step is distributing, then it is solved like any other two-step equation.
\[\begin{array}{rl}4(2y-6)=16&\text{Distribute} \\ 8y-24=16&\text{Add the opposite of }-24\text{ to each side} \\ 8y-24+\color{blue}{24}\color{black}{}=16+\color{blue}{24}\color{black}{}&\text{Simplify} \\ 8y=40&\text{Multiply by the reciprocal of }8 \\ \color{blue}{\frac{1}{8}}\color{black}{}\cdot 8y=40\cdot\color{blue}{\frac{1}{8}}\color{black}{}&\text{Simplify} \\ y=5&\text{Solution}\end{array}\nonumber\]
Solve for \(p\): \(3(2p − 4) + 9 = 15\)
Solution
Notice the first step is distributing, then it is solved like any other two-step equation.
\[\begin{array}{rl}3(2p-4)+9=15&\text{Distribute} \\ 6p-12+9=15&\text{Combine like terms} \\ 6p-3=15&\text{Add the opposite of }-3\text{ to each side} \\ 6p-3+\color{blue}{3}\color{black}{}=15+\color{blue}{3}\color{black}{}&\text{Simplify} \\ 6p=18&\text{Multiply by the reciprocal of }6 \\ \color{blue}{\frac{1}{6}}\color{black}{}\cdot 6p=18\cdot\color{blue}{\frac{1}{6}}\color{black}{}&\text{Simplify} \\ p=3&\text{Solution}\end{array}\nonumber\]
Solve for \(a\): \(3(4a − 5) = 4(2a + 1) + 5\)
Solution
Notice we have variable terms on each side of the equation. We will not only distribute first, but also isolate the variable term before applying the two-step method: \[\begin{array}{rl} 3(4a − 5) = 4(2a + 1) + 5 &\text{Distribute} \\ 12a-15=8a+4+5&\text{Combine like terms} \\ 12a-15=8a+9&\text{Isolate the variable term by adding the opposite of }8a \\ &\text{to each side} \\ 12a-15+\color{blue}{(-8a)}\color{black}{}=8a+9+\color{blue}{(-8a)}\color{black}{}&\text{Simplify} \\ 4a-15=9&\text{Add the opposite of }-15\text{ to each side} \\ 4a-15+\color{blue}{15}\color{black}{}=9+\color{blue}{15}\color{black}{}&\text{Simplify} \\ 4a=24&\text{Multiply by the reciprocal of }4 \\ \color{blue}{\frac{1}{4}}\color{black}{}\cdot 4a=24\cdot\color{blue}{\frac{1}{4}}\color{black}{}&\text{Simplify} \\ a=6&\text{Solution}\end{array}\nonumber\]
In general, there is a 5-step process to solving any linear equation. While all five steps aren’t always needed, this can serve as a guide for solving equations.
Step 1. Apply the distributive property to rewrite the equation without parenthesis.
Step 2. Combine like terms on each side of the equation.
Step 3. Isolate the variable term by applying the addition property.
Step 4. Solve the equation by applying the multiplication property.
Step 5. \(✓\) Verify the solution.
The Chinese developed a method for solving equations that involved finding each digit one at a time about 2,000 years ago.
There are two special cases when solving linear equations. The first is illustrated in the next two examples. Notice we start by distributing and moving the variables all to the same side.
Solve for \(y\): \(3(2y − 5) = 6y − 15\)
Solution
\[\begin{array}{rl} 3(2y − 5) = 6y − 15 &\text{Distribute} \\ 6y-15=6y-15&\text{Isolate the variable term by adding the opposite of }6y \\ &\text{to each side} \\ 6y-15+\color{blue}{(-6y)}\color{black}{}=6y+15+\color{blue}{(-6y)}\color{black}{}&\text{Simplify} \\ -15\stackrel{?}{=}-15&\text{Is this true? }\color{blue}{\text{Yes }\checkmark} \color{black}{} \\ -15=-15\end{array}\nonumber\]
Notice all variables canceled and we are left with only a statement with numbers. In this case, the statement we are left with is a true statement, i.e., \(−15\) does equal \(−15\), and so there are infinitely many solutions to this equation. In this special case, when we obtain infinitely many solutions, then the solution is all real numbers. When the solution is all real numbers, we call this equation an identity.
Solve for \(n\): \(2(3n − 5) − 4n = 2n + 7\)
Solution
\[\begin{array}{rl} 2(3n-5)-4n=2n+7&\text{Distribute} \\ 6n-10-4n=2n+7&\text{Combine like terms} \\ 2n-10=2n+7&\text{Isolate the variable term by adding the opposite of }2n \\ &\text{to each side} \\ -10\stackrel{?}{=}7 &\text{Is this true?}\color{blue}{\text{ No }X} \color{black}{} \\ -10\neq 7\end{array}\nonumber\]
Notice all variables canceled and we are left with only a statement with numbers. In this case, the statement we are left with is a false statement, i.e., \(−10\) does not equal \(7\), and so there is no solution to this equation. In this special case, when we obtain a false statement, then the solution is no solution and we call this equation a contradiction.
There are three types of equations we saw in the examples and in Examples \(\PageIndex{24}\) and \(\PageIndex{25}\):
- An equation is a conditional equation when there is one unique solution, i.e., \(x =\text{ number}\).
- An equation is an identity when we obtain infinitely many solutions, as in Example \(\PageIndex{24}\), where the solution to the equation is all real numbers. Hence, we can choose any number for the variable and this number will make the statement true.
- An equation is a contradiction when the solution is no solution, as in Example \(\PageIndex{25}\). Hence, there is no number for the variable such that the statement is true.
Solving Equations with Fractions
Often when solving linear equations, we work with an equation with fraction coefficients. We can solve these problems as usual. Recall, the methods never change, just problems.
Solve for \(y\): \(\frac{3}{4}y-\frac{7}{2}=\frac{5}{6}\)
Solution
\[\begin{array}{rl}\frac{3}{4}y-\frac{7}{2}=\frac{5}{6}&\text{Isolate the variable term by adding the opposite of }-\frac{7}{2} \\ \frac{3}{4}y-\frac{7}{2}+\color{blue}{\frac{7}{2}}\color{black}{}=\frac{5}{6}+\color{blue}{\frac{7}{2}}\color{black}{}&\text{Simplify}\end{array}\nonumber\]
Notice, in order to add \(\frac{5}{6} + \frac{7}{2}\), we need to obtain the LCD. Since the \(\text{LCD}(2, 6) = 6\), we can rewrite the right side with the LCD: \[\begin{array}{rl}\frac{3}{4}y=\frac{5}{6}+\frac{21}{6}&\text{Combine like terms} \\ \frac{3}{4}y=\frac{26}{6}&\text{Simplify }\frac{26}{6} \\ \frac{3}{4}y=\frac{13}{3}&\text{Solve by multiplying by the reciprocal of }\frac{3}{4} \\ \color{blue}{\frac{4}{3}}\color{black}{}\cdot\frac{3}{4}y=\frac{13}{3}\cdot\color{blue}{\frac{4}{3}}\color{black}{}&\text{Simplify} \\ y=\frac{52}{9}&\text{Solution}\end{array}\nonumber\]
While this process does help us arrive at the correct solution, the fractions can make the process quite difficult and we are more inclined to make errors. Hence, we have an alternate method called clearing denominators.
We can easily clear denominators in an equation by multiplying each term by the LCD. After completing this step, the fractions are cleared and we can work with a more familiar type of equation.
Let’s try Example \(\PageIndex{26}\) again, but, now, by clearing denominators first, then solving.
Solve for \(y\): \(\frac{3}{4}y-\frac{7}{2}=\frac{5}{6}\)
Solution
\[\begin{array}{rl}\frac{3}{4}y-\frac{7}{2}=\frac{5}{6}&\text{Multiply each term by the LCD}(2,4,6)=12 \\ \color{blue}{12}\color{black}{}\cdot\frac{3}{4}y-\color{blue}{12}\color{black}{}\cdot\frac{7}{2}=\color{blue}{12}\color{black}{}\cdot\frac{5}{6}&\text{Simplify} \\ \frac{36}{4}y-\frac{84}{2}=\frac{60}{6}&\text{Reduce the fractions} \\ 9y-42=10&\text{Add the opposite of }-42\text{ to each side} \\ 9y-42+\color{blue}{42}\color{black}{}=10+\color{blue}{42}\color{black}{}&\text{Simplify} \\ 9y=52&\text{Multiply by the reciprocal of }9 \\ \color{blue}{\frac{1}{9}}\color{black}{}\cdot 9y=52\cdot\color{blue}{\frac{1}{9}}\color{black}{}&\text{Simplify} \\ y=\frac{52}{9}&\text{Solution}\end{array}\nonumber\]
Thus, the solution is \(y = \frac{52}{9}\) and Example \(\PageIndex{26}\) is a conditional equation. Also, we observe that as soon as we multiplied each term by the LCD, we cleared the denominators and the equation no longer contained fractions.
Solve for \(t\): \(\frac{2}{3}t-2=\frac{3}{2}t+\frac{1}{6}\)
Solution
\[\begin{array}{rl}\frac{2}{3}t-2=\frac{3}{2}t+\frac{1}{6}&\text{Multiply each term by the LCD}(2,3,6)=6 \\ \color{blue}{6}\color{black}{}\cdot\frac{2}{3}t-\color{blue}{6}\color{black}{}\cdot 2=\color{blue}{6}\color{black}{}\cdot\frac{3}{2}t+\color{blue}{6}\color{black}{}\cdot\frac{1}{6}&\text{Simplify} \\ \frac{12}{3}t-12=\frac{18}{2}t+\frac{6}{6}&\text{Reduce the fractions} \\ 4t-12=9t+1&\text{Isolate the variable term by adding the opposite of }9t \\ &\text{to each side} \\ 4t-12+\color{blue}{(-9t)}\color{black}{}=9t+1+\color{blue}{(-9t)}\color{black}{}&\text{Simplify} \\ -5t-12=1&\text{Add the opposite of }-12\text{ to each side} \\ -5t-12+\color{blue}{12}\color{black}{}=1+\color{blue}{12}\color{black}{}&\text{Simplify} \\ -5t=13&\text{Multiply by the reciprocal of }-5 \\ \color{blue}{-\frac{1}{5}}\color{black}{}\cdot -5t=13\cdot\color{blue}{-\frac{1}{5}}\color{black}{}&\text{Simplify} \\ t=-\frac{13}{5}&\text{Solution}\end{array}\nonumber\]
Thus, the solution is \(t = − \frac{13}{5}\) and Example \(\PageIndex{28}\) is a conditional equation.
In Example \(\PageIndex{28}\), we could write the solution as \(−2.6\) given we were using a calculator. A good rule of thumb is if you start with fractions, the solution should also be a fraction. Hence, since there were fractions in original problem, we will leave the solution as a fraction.
The Egyptians were among the first to study fractions and linear equations. The most famous mathematical document from Ancient Egypt is the Rhind Papyrus, where the unknown variable was called “heap.”
Solving Equations with Distributing Fractions
We can use this same method if parenthesis are in the given problem. We will first distribute the coefficient in front of the parenthesis, then clear denominators.
Solve for \(y\): \(\frac{3}{2}\left(\frac{5}{9}y+\frac{4}{27}\right)=3\)
Solution
\[\begin{array}{rl}\frac{3}{2}\left(\frac{5}{9}y+\frac{4}{27}\right)=3&\text{Distribute }\frac{3}{2}\text{ and reduce} \\ \frac{5}{6}y+\frac{2}{9}=3&\text{Multiply each term by the LCD}(6,9)=18 \\ \color{blue}{18}\color{black}{}\cdot\frac{5}{6}y+\color{blue}{18}\color{black}{}\cdot\frac{2}{9}=\color{blue}{18}\color{black}{}\cdot 3&\text{Multiply and simplify} \\ 15y+4=54&\text{Add the opposite of }4\text{ to each side} \\ 15y+4+\color{blue}{(-4)}\color{black}{}=54+\color{blue}{(-4)}\color{black}{}&\text{Simplify} \\ 15y=50&\text{Multiply by the reciprocal of }15 \\ \color{blue}{\frac{1}{15}}\color{black}{}\cdot 15y=50\cdot\color{blue}{\frac{1}{15}}\color{black}{}&\text{Simplify} \\ y=\frac{50}{15}&\text{Reduce} \\ y=\frac{10}{3}&\text{Solution}\end{array}\nonumber\]
Thus, the solution is \(y = \frac{10}{3}\) and Example \(\PageIndex{29}\) is a conditional equation.
Solve for \(q\): \(\frac{1}{4}q-\frac{1}{2}=\frac{1}{3}\left(\frac{3}{4}q+6\right)-\frac{7}{2}\)
Solution
\[\begin{array}{rl}\frac{1}{4}q-\frac{1}{2}=\frac{1}{3}\left(\frac{3}{4}q+6\right)-\frac{7}{2}&\text{Distribute }\frac{1}{3}\text{ and reduce} \\ \frac{1}{4}q-\frac{1}{2}=\frac{1}{4}q+2-\frac{7}{2}&\text{Multiply each term by the LCD}(2,4)=4 \\ \color{blue}{4}\color{black}{}\cdot\frac{1}{4}q-\color{blue}{4}\color{black}{}\cdot\frac{1}{2}=\color{blue}{4}\color{black}{}\cdot\frac{1}{4}q+\color{blue}{4}\color{black}{}\cdot 2-\color{blue}{4}\color{black}{}\cdot\frac{7}{2}&\text{Multiply and reduce} \\ q-2=q+8-14&\text{Combine like terms} \\ q-2=q-6&\text{Isolate the variable term by adding the opposite of }q \\ &\text{to each side} \\ q-2+\color{blue}{(-q)}\color{black}{}=q-6+\color{blue}{(-q)}\color{black}{}&\text{Simplify} \\ -2\stackrel{?}{=}-6&\text{Is this true? }\color{blue}{\text{No }X}\color{black}{} \\ -2\neq -6&\text{This implies there is no solution}\end{array}\nonumber\]
Since we obtain a false statement, there is no solution and this equation is called a contradiction.
Linear Equations Homework
Solve the one-step equations.
\(v + 9 = 16\)
\(x − 11 = −16\)
\(30 = a + 20\)
\(x − 7 = −26\)
\(13 = n − 5\)
\(340 = −17x\)
\(−9 = \frac{n}{12}\)
\(20v = −160\)
\(340 = 20n\)
\(16x = 320\)
\(−16 + n = −13\)
\(p − 8 = −21\)
\(180 = 12x\)
\(20b = −200\)
\(\frac{r}{14}=\frac{5}{14}\)
\(−7 = a + 4\)
\(10 = x − 4\)
\(13a = −143\)
\(\frac{p}{20} = −12\)
\(9 + m = −7\)
\(14 = b + 3\)
\(−14 = x − 18\)
\(−1 + k = 5\)
\(−13 + p = −19\)
\(22 = 16 + m\)
\(4r = −28\)
\(\frac{5}{9} = \frac{b}{9}\)
\(−20x = −80\)
\(\frac{1}{2} = \frac{a}{8}\)
\(\frac{k}{13}= −16\)
\(21 = x + 5\)
\(m − 4 = −13\)
\(3n = 24\)
\(−17 = \frac{x}{12}\)
\(n + 8 = 10\)
\(v − 16 = −30\)
\(−15 = x − 16\)
\(-8k=120\)
\(-15=\frac{x}{9}\)
\(-19=\frac{n}{20}\)
Solve the two-step equations.
\(5 + \frac{n}{4}= 4\)
\(102 = −7r + 4\)
\(−8n + 3 = −77\)
\(0 = −6v\)
\(−8 = \frac{x}{5}− 6\)
\(0 = −7 + \frac{k}{2}\)
\(−12 + 3x = 0\)
\(24 = 2n − 8\)
\(2 = −12 + 2r\)
\(\frac{b}{3} + 7 = 10\)
\(152 = 8n + 64\)
\(−16 = 8a + 64\)
\(56 + 8k = 64\)
\(−2x + 4 = 22\)
\(−20 = 4p + 4\)
\(−5 = 3 + \frac{n}{2}\)
\(\frac{r}{8} − 6 = −5\)
\(−40 = 4n − 32\)
\(87 = 3 − 7v\)
\(−x + 1 = −11\)
\(−2 = −2m + 12\)
\(27 = 21 − 3x\)
\(−4 − b = 8\)
\(−2 + \frac{x}{2} = 4\)
\(−5 = \frac{a}{4} − 1\)
\(−6 = 15 + 3p\)
\(−5m + 2 = 27\)
\(−37 = 8 + 3x\)
\(−8 + \frac{n}{12} = −7\)
\(\frac{x}{1} − 8 = −8\)
\(−11 = −8 + \frac{v}{2}\)
\(−2x − 3 = −29\)
\(−4 − 3n = −16\)
\(67 = 5m − 8\)
\(9 = 8 + \frac{x}{6}\)
\(\frac{m}{4} − 1 = −2\)
\(−80 = 4x − 28\)
\(33 = 3b + 3\)
\(3x − 3 = −3\)
\(4 + \frac{a}{3} = 1\)
Solve.
\(2 − (−3a − 8) = 1\)
\(−5 (−4 + 2v) = −50\)
\(66 = 6 (6 + 5x)\)
\(0 = −8 (p − 5)\)
\(−2 + 2 (8x − 7) = −16\)
\(−21x + 12 = −6 − 3x\)
\(−1 − 7m = −8m + 7\)
\(1 − 12r = 29 − 8r\)
\(20 − 7b = −12b + 30\)
\(−32 − 24v = 34 − 2v\)
\(−2 − 5 (2 − 4m) = 33 + 5m\)
\(−4n + 11 = 2 (1 − 8n) + 3n\)
\(−6v − 29 = −4v − 5 (v + 1)\)
\(2 (4x − 4) = −20 − 4x\)
\(−a − 5 (8a − 1) = 39 − 7a\)
\(−57 = − (−p + 1) + 2 (6 + 8p)\)
\(−2 (m − 2) + 7 (m − 8) = −67\)
\(50 = 8 (7 + 7r) − (4r + 6)\)
\(−8 (n − 7) + 3 (3n − 3) = 41\)
\(−61 = −5 (5r − 4) + 4 (3r − 4)\)
\(−2 (8n − 4) = 8 (1 − n)\)
\(−3 (−7v + 3) + 8v = 5v − 4 (1 − 6v)\)
\(−7 (x − 2) = −4 − 6 (x − 1)\)
\(−6 (8k + 4) = −8 (6k + 3) − 2\)
\(−2 (1 − 7p) = 8 (p − 7)\)
\(2 (−3n + 8) = −20\)
\(2 − 8 (−4 + 3x) = 34\)
\(32 = 2 − 5 (−4n + 6)\)
\(−55 = 8 + 7 (k − 5)\)
\(− (3 − 5n) = 12\)
\(−3n − 27 = −27 − 3n\)
\(56p − 48 = 6p + 2\)
\(4 + 3x = −12x + 4\)
\(−16n + 12 = 39 − 7n\)
\(17 − 2x = 35 − 8x\)
\(−25 − 7x = 6 (2x − 1)\)
\(−7 (1 + b) = −5 − 5b\)
\(−8 (8r − 2) = 3r + 16\)
\(−8n − 19 = −2 (8n − 3) + 3n\)
\(−4 + 4k = 4 (8k − 8)\)
\(16 = −5 (1 − 6x) + 3 (6x + 7)\)
\(7 = 4 (n − 7) + 5 (7n + 7)\)
\(−8 (6 + 6x) + 4 (−3 + 6x) = −12\)
\(−76 = 5 (1 + 3b) + 3 (3b − 3)\)
\(−6 (x − 8) − 4 (x − 2) = −4\)
\(−4 (1 + a) = 2a − 8 (5 + 3a)\)
\(−6 (x − 3) + 5 = −2 − 5 (x − 5)\)
\(− (n + 8) + n = −8n + 2 (4n − 4)\)
\(−5 (x + 7) = 4 (−8x − 2)\)
\(8 (−8n + 4) = 4 (−7n + 8)\)
Solve.
\(\frac{3}{5}(1+p)=\frac{21}{20}\)
\(0=-\frac{5}{4}\left(x-\frac{6}{5}\right)\)
\(\frac{3}{4}-\frac{5}{4}m=\frac{113}{24}\)
\(\frac{635}{72}=-\frac{5}{2}\left(-\frac{11}{4}+x\right)\)
\(2b+\frac{9}{5}=-\frac{11}{5}\)
\(\frac{3}{2}\left(\frac{7}{3}n+1\right)=\frac{3}{2}\)
\(-a-\frac{5}{4}\left(-\frac{8}{3}a+1\right)=-\frac{19}{4}\)
\(\frac{55}{6}=-\frac{5}{2}\left(\frac{3}{2}p-\frac{5}{3}\right)\)
\(\frac{16}{9}=-\frac{4}{3}\left(-\frac{4}{3}n-\frac{4}{3}\right)\)
\(-\frac{5}{8}=\frac{5}{4}\left(r-\frac{3}{2}\right)\)
\(-\frac{11}{3}+\frac{3}{2}b=\frac{5}{2}\left(b-\frac{5}{3}\right)\)
\(-\left(-\frac{5}{2}x-\frac{3}{2}\right)=-\frac{3}{2}+x\)
\(\frac{45}{16}+\frac{3}{2}n=\frac{7}{4}n-\frac{19}{16}\)
\(\frac{3}{2}\left(v+\frac{3}{2}\right)=-\frac{7}{4}v-\frac{19}{6}\)
\(\frac{47}{9}+\frac{3}{2}x=\frac{5}{3}\left(\frac{5}{2}x+1\right)\)
\(-\frac{1}{2}=\frac{3}{2}k+\frac{3}{2}\)
\(\frac{3}{2}n-\frac{8}{3}=-\frac{29}{12}\)
\(\frac{11}{4}+\frac{3}{4}r=\frac{163}{32}\)
\(-\frac{16}{9}=-\frac{4}{3}\left(\frac{5}{3}+n\right)\)
\(\frac{3}{2}-\frac{7}{4}v=-\frac{9}{8}\)
\(\frac{41}{9}=\frac{5}{2}\left(x+\frac{2}{3}\right)-\frac{1}{3}x\)
\(\frac{1}{3}\left(-\frac{7}{4}k+1\right)-\frac{10}{3}k=-\frac{13}{8}\)
\(-\frac{1}{2}\left(\frac{2}{3}x-\frac{3}{4}\right)-\frac{7}{2}x=-\frac{83}{24}\)
\(\frac{2}{3}\left(m+\frac{9}{4}\right)-\frac{10}{3}=-\frac{53}{18}\)
\(\frac{1}{12}=\frac{4}{3}x+\frac{5}{3}\left(x-\frac{7}{4}\right)\)
\(\frac{7}{6}-\frac{4}{3}n=-\frac{3}{2}n+2\left(n+\frac{3}{2}\right)\)
\(-\frac{149}{16}-\frac{11}{3}r=-\frac{7}{4}r-\frac{5}{4}\left(-\frac{4}{3}r+1\right)\)
\(-\frac{7}{2}\left(\frac{5}{3}a+\frac{1}{3}\right)=\frac{11}{4}a+\frac{25}{8}\)
\(-\frac{8}{3}-\frac{1}{2}x=-\frac{4}{3}x-\frac{2}{3}\left(-\frac{13}{4}x+1\right)\)
\(\frac{1}{3}n+\frac{29}{6}=2\left(\frac{4}{3}n+\frac{2}{3}\right)\)