Solve the given linear equation, then check your answer − ( h + 3 ) = 11 -\left(h+3\right)=11

h = − 11 3 h=-\frac{11}{3} h = − 3 11

2. Linear Equations and Inequalities

Solving Linear Equations

2. Linear Equations and Inequalities

Solving Linear Equations: Videos & Practice Problems

Topic summary

Linear equations, expressed as $a x + b = c$ , involve isolating the variable by applying properties of equality: addition, subtraction, multiplication, and division. These operations maintain equation balance, akin to a scale, ensuring equivalent equations. Fractional coefficients are handled by multiplying both sides by the reciprocal, simplifying the variable's isolation. Solving involves simplifying expressions, distributing terms, combining like terms, and systematically isolating the variable. Checking solutions by substitution confirms accuracy. Understanding coefficients, terms, and operations is essential for mastering linear equations and polynomial expressions in standard form.

concept

Introduction to Linear Equations

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Introduction to Linear Equations Video Summary

An equation is a mathematical statement asserting that two algebraic expressions are equal. A specific type of equation, known as a linear equation, can be expressed in the form $ax + b = c$, where $a$, $b$, and $c$ are real numbers and $a \neq 0$. This condition ensures the presence of a variable term, which is essential for the equation to be linear. The variable, commonly denoted as $x$, can actually be any letter, and the term "linear" or "first degree equation" refers to the fact that the variable is raised to the power of one, indicating the equation's degree.

The solution to a linear equation is the value of the variable that makes the equation true when substituted back into it. For example, consider the equation \$2x + 6 = 0$. To determine if $x = -3$ is a solution, substitute $-3$ for $x\(:

\[2(-3) + 6 = -6 + 6 = 0\]

Since the left side equals the right side, \)x = -3$ is indeed a solution. Conversely, for the equation $5 = 8w - 3$, substituting $w = -1\( yields:

\[8(-1) - 3 = -8 - 3 = -11\]

Since \)5 \neq -11$, $w = -1\( is not a solution. This illustrates the process of verifying solutions by substitution, a fundamental skill in algebra.

Sometimes, linear equations may not initially appear in the standard form \)ax + b = c$, but they can be rearranged to fit this format. For instance, $5 = 8w - 3$ can be rewritten as $8w - 3 = 5$, clearly showing $a = 8$, $b = -3$, and $c = 5\(.

The set of all solutions to a linear equation is called the solution set, typically denoted using curly brackets. For example, if \)x = -3$ is the only solution to $2x + 6 = 0$, the solution set is written as $\{-3\}$. Understanding how to identify and express solution sets is crucial for solving and interpreting linear equations effectively.

example

Introduction to Linear Equations Example 1

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Introduction to Linear Equations Example 1 Video Summary

To determine whether an expression is a linear equation in one variable, it must first be an equation, meaning it has an equal sign separating two expressions. For example, an expression like 4x - 7 + 3 is not an equation because it lacks an equal sign, so it cannot be classified as a linear equation in one variable.

When examining an equation such as 5y - y = 2, it qualifies as an equation since it has two expressions separated by an equal sign. Although the variable y appears twice, both instances are to the first power, which is characteristic of linear equations. Combining like terms simplifies this to 4y = 2, clearly showing it is a linear equation in one variable.

In contrast, an equation like 6x + 1 = 2t² involves two different variables, x and t, and the variable t is squared. Linear equations in one variable must contain only one variable raised to the first power. Therefore, this is not a linear equation in one variable.

In summary, a linear equation in one variable must have exactly one variable, which appears only to the first power, and must be presented as an equation with an equal sign. Expressions without an equal sign or with multiple variables or variables raised to powers other than one do not meet the criteria for linear equations in one variable.

Problem

Identify the following as either an expression or equation.
$\frac{2 m}{3} + 8$

Expression

Equation

Problem

Identify the following as either an expression or equation.
$4 (a - 2) = 21$

Expression

Equation

Problem

Which of the following is a linear equation in one variable?

$x+5=12$

$x^2=25$

$y+z=10$

$x-3<7$

Problem

Which of the following is a linear equation in one variable?

$2\left(x+5\right)-3x=x^2+1$

$x\left(x+4\right)=0$

$4\left(x+3\right)=2x+18$

$x^2+5x=10$

Problem

Verify that the given value is a solution to the equation.
$y = - 2; 5 y + 4 = 14$

Solution

Not a solution

Problem

Verify that the given value is a solution to the equation.
$a = 2; 4 a + 3 = a + 9$

Solution

Not a solution

concept

Addition and Subtraction Properties of Equality

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Addition and Subtraction Properties of Equality Video Summary

A linear equation is an algebraic statement where the variable is to the first power, and solving it means finding the value of the variable that makes the equation true. For example, in the equation x − 6 = 0, the solution is the value of x that satisfies this equality.

To solve linear equations, the primary goal is to isolate the variable on one side of the equation. This is achieved by performing operations such as addition or subtraction on both sides of the equation, ensuring the creation of equivalent equations. Equivalent equations have the same solution because whatever operation is done to one side must be done to the other to maintain balance, much like a scale where adding or removing equal weights on both sides keeps it balanced.

The addition property of equality states that if a = b, then adding the same value c to both sides results in a + c = b + c. This property is especially useful when an equation involves subtraction, as adding the opposite value cancels out the subtraction. For instance, in x − 6 = 0, adding 6 to both sides cancels the −6, isolating x and yielding x = 6. This solution can be verified by substituting back into the original equation: 6 − 6 = 0, which is true.

Similarly, the subtraction property of equality states that if a = b, then subtracting the same value c from both sides results in a − c = b − c. This property is used when an equation has addition, allowing subtraction to cancel it out. For example, in 0 = x + 2, subtracting 2 from both sides isolates x, giving x = −2. Checking the solution by substitution confirms its correctness.

When solving linear equations, the solution should always isolate the variable, either as x = number or number = x. The key is that the variable stands alone on one side. By applying the addition and subtraction properties of equality appropriately, one can systematically solve linear equations by canceling out unwanted terms and isolating the variable.

example

Addition and Subtraction Properties of Equality Example 2

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Addition and Subtraction Properties of Equality Example 2 Video Summary

To solve the linear equation y − 1.2 = 5.8, the goal is to isolate the variable y on one side. Since 1.2 is being subtracted from y, you can eliminate this by adding 1.2 to both sides of the equation. This maintains the equality because whatever operation you perform on one side must be done on the other side as well.

Instead of rewriting the entire equation horizontally after adding 1.2, you can perform the addition vertically for efficiency. Adding 1.2 to −1.2 cancels out the subtraction, leaving y alone on the left side. On the right side, adding 1.2 to 5.8 results in 7.0. Thus, the solution is:

\[ y = 7 \]

To verify this solution, substitute y = 7 back into the original equation:

\[ 7 - 1.2 = 5.8 \]

Since the left side equals the right side, the solution is correct. This process highlights the importance of performing inverse operations to isolate variables and checking solutions by substitution to confirm their validity. Using vertical addition or subtraction can streamline solving linear equations without altering the mathematical integrity of the process.

Problem

Solve the given linear equation using addition and subtraction properties of equality.
$x + \frac{2}{8} = - \frac{3}{8}$

$x=\frac58$

$x=-\frac58$

$x=5$

$x=-5$

Problem

Solve the given linear equation using addition and subtraction properties of equality.
$- 5.4 + c = 1.6$

$c=1.6$

$c=-1.6$

$c=7.0$

$c=-7.0$

Problem

Solve the given linear equation using addition and subtraction properties of equality.
$6 h - (- 12) = 5 + 5 h$

$h=-17$

$h=17$

$h=7$

$h=-7$

Problem

Solve the given linear equation using addition and subtraction properties of equality.
$2 (x + 5) = 3 (x - 1)$

$x=-13$

$x=13$

$x=3$

$x=-3$

Problem

Solve the given linear equation using addition and subtraction properties of equality.
$3 (y + 3) + (1 - y) = 3 y + 13$

$y=-4$

$y=4$

$y=0$

$y=13$

example

Addition and Subtraction Properties of Equality Example 3

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Addition and Subtraction Properties of Equality Example 3 Video Summary

To translate the statement "a number decreased by seven is equal to 15" into a linear equation, start by representing the unknown number with a variable, commonly x. The phrase "decreased by seven" indicates subtraction, so the expression becomes x - 7. Since this is equal to 15, the equation is written as:

\[x - 7 = 15\]

To solve for x, isolate the variable by undoing the subtraction. Add 7 to both sides of the equation to maintain balance, resulting in:

\[x - 7 + 7 = 15 + 7\]

which simplifies to

\[x = 22\]

This means the number is 22. To verify the solution, substitute 22 back into the original equation:

\[22 - 7 = 15\]

Since this is a true statement, the solution is confirmed. This process demonstrates how to translate verbal expressions into linear equations and solve them by applying inverse operations, a fundamental skill in algebra.

concept

Multiplication and Division Properties of Equality

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Multiplication and Division Properties of Equality Video Summary

When solving linear equations, it is essential to maintain balance by performing the same operation on both sides of the equation. This principle ensures that the equality remains true throughout the process. Beyond addition and subtraction, multiplication and division properties of equality are powerful tools used to isolate variables and solve equations effectively.

The multiplication property of equality states that if two expressions are equal, multiplying both sides by the same nonzero number preserves the equality. Formally, if $a = b$, then $a \times c = b \times c$. This property is particularly useful when an equation involves division, as multiplication can cancel out the division to isolate the variable. For example, in the equation $\frac{x}{2} = 9$, multiplying both sides by 2 eliminates the denominator, resulting in $x = 18$.

Conversely, the division property of equality asserts that dividing both sides of an equation by the same nonzero number maintains equality. If $a = b$, then $\frac{a}{c} = \frac{b}{c}$. This property is especially helpful when the variable is multiplied by a coefficient. For instance, in \$20 = 5x$, dividing both sides by 5 isolates $x$, yielding $x = 4\(.

When solving more complex linear equations, it is important to first use addition or subtraction to isolate the variable term before applying multiplication or division to solve for the variable itself. Consider the equation \)3a - 4 = 11$. To isolate the term with the variable, add 4 to both sides, resulting in $3a = 15$. Then, divide both sides by 3 to solve for $a$, giving $a = 5\(.

Verifying solutions by substituting the found value back into the original equation confirms the correctness of the solution. For example, substituting \)a = 5$ into $3a - 4 = 11$ yields $3 \times 5 - 4 = 15 - 4 = 11$, confirming the solution is accurate.

Mastering these properties of equality—addition, subtraction, multiplication, and division—enables the systematic solving of linear equations by isolating variables step-by-step while maintaining equation balance. This foundational skill is critical for progressing to more advanced algebraic concepts.

example

Multiplication and Division Properties of Equality Example 1

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Multiplication and Division Properties of Equality Example 1 Video Summary

To solve the linear equation 1.2a + 2.3 = 5.9, the goal is to isolate the variable a. Begin by using the subtraction property of equality to eliminate the constant term on the left side. Subtract 2.3 from both sides to maintain balance, resulting in 1.2a = 5.9 - 2.3. Simplifying the right side gives 1.2a = 3.6. Next, apply the division property of equality by dividing both sides by 1.2 to solve for a. This yields a = \frac{3.6}{1.2} = 3.

To verify the solution, substitute a = 3 back into the original equation. Calculate 1.2 \times 3 + 2.3, which equals 3.6 + 2.3 = 5.9. Since both sides of the equation are equal, the solution is confirmed as correct.

This process highlights the importance of using properties of equality—subtraction and division—to isolate variables in linear equations. Understanding how to manipulate equations while maintaining equality is fundamental in algebra. The solution method ensures that the variable is isolated step-by-step, and checking the solution reinforces accuracy and comprehension.

Problem

Solve the given linear equation using multiplication and division properties of equality.
$- 8 x = 64$

$x=64$

$x=-64$

$x=-8$

$x=8$

Problem

Solve the given linear equation using multiplication and division properties of equality.
$\frac{y}{4} = - \frac{21}{6}$

$y=14$

$y=-14$

$y=7$

$y=-7$

Problem

Solve the given linear equation using multiplication and division properties of equality.
$- \frac{126}{14} = 3 y$

$y=27$

$y=-27$

$y=3$

$y=-3$

Problem

Solve the given linear equation, then check your answer
$- (h + 3) = 11$

$h=-14$

$h=-\frac{11}{3}$

$h=-33$

$h=33$

Problem

Solve the given linear equation, then check your answer
$0.5 t + 1.5 t = 7 + 3 - 4$

$t=1$

$t=-6$

$t=3$

$t=-3$

example

Multiplication and Division Properties of Equality Example 2

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Multiplication and Division Properties of Equality Example 2 Video Summary

To translate the statement "three times L equals 54" into a linear equation, recognize that "three times" indicates multiplication. This means the expression can be written as \$3L = 54$, where $L$ is the variable representing the unknown value. The goal is to solve for $L\(.

Since \)3$ is multiplied by $L$, apply the division property of equality to isolate $L$. Divide both sides of the equation by $3\( to maintain balance:

\[3L = 54\]

\[\frac{3L}{3} = \frac{54}{3}\]

\[L = 18\]

This shows that \)L$ equals $18$. To verify the solution, substitute $18\( back into the original equation:

\[3 \times 18 = 54\]

Since \)54 = 54$ is a true statement, the solution $L = 18$ is correct. This process demonstrates how to translate verbal expressions into algebraic equations and solve them using inverse operations, reinforcing the fundamental concept of maintaining equality while isolating variables.

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Multiplication Property for Fraction Coefficients

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Multiplication Property for Fraction Coefficients Video Summary

When solving linear equations where the variable has a fractional coefficient, such as $\frac{3}{4}x = 9$, the key strategy is to eliminate the fraction by multiplying both sides of the equation by the reciprocal of the fraction. The reciprocal of a fraction is obtained by flipping its numerator and denominator. For example, the reciprocal of $\frac{3}{4}$ is $\frac{4}{3}$. This works because multiplying a number by its reciprocal always results in 1, as shown by the equation $\frac{2}{3} \times \frac{3}{2} = 1$.

Applying this to the equation $\frac{3}{4}x = 9$, multiply both sides by $\frac{4}{3}$ to isolate $x$:

\[\frac{4}{3} \times \frac{3}{4} x = \frac{4}{3} \times 9\]

The left side simplifies to \$1 \times x = x\(, and the right side becomes:

\[x = 9 \times \frac{4}{3} = \frac{9}{1} \times \frac{4}{3} = \frac{36}{3} = 12\]

Thus, \)x = 12\(. This method ensures the variable is isolated by effectively "canceling out" the fractional coefficient.

Consider another example: \)10 = \frac{5}{3} y$. To solve for $y$, multiply both sides by the reciprocal of $\frac{5}{3}$, which is $\frac{3}{5}\(:

\[\frac{3}{5} \times 10 = \frac{3}{5} \times \frac{5}{3} y\]

Simplifying the right side yields \)1 \times y = y\(, and the left side becomes:

\[y = \frac{3}{5} \times \frac{10}{1} = \frac{30}{5} = 6\]

Therefore, \)y = 6$. This approach leverages the multiplication property of equality, which states that multiplying both sides of an equation by the same nonzero number maintains equality, allowing for the elimination of fractional coefficients.

By mastering the use of reciprocals to clear fractional coefficients, solving linear equations becomes more straightforward and efficient. This technique is essential for handling a wide range of linear equations and ensures that variables can be isolated cleanly, facilitating further algebraic manipulation and problem-solving.

Problem

Solve the given linear equation, then check your answer.
$\frac{1}{2} = \frac{3}{4} x$

$x=\frac38$

$x=2$

$x=\frac23$

$x=\frac13$

concept

Strategy for Solving Linear Equations

Video duration:

Strategy for Solving Linear Equations Video Summary

Solving linear equations involves a systematic approach that combines simplification and the application of properties of equality to isolate the variable. The process begins by simplifying both sides of the equation, which includes distributing any coefficients across parentheses and combining like terms. For example, when given an equation such as \$3(x - 2) + 2 = x + 8$, start by distributing the 3 to both $x$ and $-2$, resulting in $3x - 6 + 2 = x + 8$. Then, combine the constants on the left side to simplify it to $3x - 4 = x + 8\(.

Next, use the addition and subtraction properties of equality to gather all variable terms on one side and constants on the other. This means subtracting \)x$ from both sides to get $3x - x - 4 = 8$, which simplifies to $2x - 4 = 8$. Then, add 4 to both sides to isolate the term with the variable, yielding $2x = 12\(.

To fully isolate the variable \)x$, apply the multiplication or division property of equality. Since $x$ is multiplied by 2, divide both sides by 2 to get $x = \frac{12}{2} = 6\(. This step ensures that the variable stands alone on one side of the equation, providing the solution.

Finally, always verify the solution by substituting the value back into the original equation. Plugging \)x = 6$ into $3(x - 2) + 2 = x + 8$ gives $3(6 - 2) + 2 = 6 + 8$. Simplifying both sides results in $3 \times 4 + 2 = 14$ and $14 = 14$, confirming the solution is correct.

This methodical approach to solving linear equations—simplifying expressions, using addition and subtraction properties to group like terms, applying multiplication or division to isolate the variable, and verifying the solution—builds a strong foundation for tackling a wide range of algebraic problems with confidence and accuracy.

Problem

Solve the given linear equation. Check your solution.
$2 (x + 3) = 14$

$x=11$

$x=4$

$x=10$

$x=5.5$

Problem

Solve the given linear equation. Check your solution.
$- 5 - y = 3 (y + 9)$

$y=-8$

$y=8$

$y=-3$

$y=3$

Problem

Solve the given linear equation. Check your solution.
$4 (x + 1) - 3 (x - 2) = 2 x + 5$

$x=3$

$x=0$

$x=-5$

$x=5$

Here’s what students ask on this topic:

A linear equation is an algebraic equation that can be written in the form $a x + b = c$ , where $a, b, c$ are real numbers and $\neq 0$ . The variable term has an exponent of one, which is why linear equations are also called first-degree equations. To identify a linear equation, check if the variable is raised only to the first power and the equation can be rearranged into the form $a x + b = c$ . For example, $2 x + 6 = 0$ is linear because it fits this form. Linear equations graph as straight lines and their solutions are values of the variable that make the equation true when substituted back in.

To solve a linear equation using the addition and subtraction properties of equality, the goal is to isolate the variable by eliminating constants on the variable's side. The addition property states that if $a = b$ , then $a + c = b + c$ . Similarly, the subtraction property states that if $a = b$ , then $a - c = b - c$ . For example, to solve $x - 6 = 0$ , add 6 to both sides to get $x = 6$ . Always perform the same operation on both sides to maintain equality. After isolating the variable, check your solution by substituting it back into the original equation to verify it makes the equation true.

The multiplication property of equality states that if $a = b$ , then $a \times c = b \times c$ . The division property states that if $a = b$ , then $a \div c = b \div c$ , where $\neq 0$ . These properties are used to isolate variables when they are multiplied or divided by a number. For example, to solve $x \div 2 = 9$ , multiply both sides by 2 to get $x = 18$ . Similarly, to solve $20 = 5 \times x$ , divide both sides by 5 to get $x = 4$ . Always perform the same multiplication or division on both sides to maintain equality and check your solution by substitution.

When a variable has a fractional coefficient, solve the linear equation by multiplying both sides by the reciprocal of the fraction. The reciprocal of a fraction is obtained by flipping its numerator and denominator. Multiplying a fraction by its reciprocal equals one, which cancels the fraction coefficient. For example, to solve $\frac{3}{4} x = 9$ , multiply both sides by the reciprocal $\frac{4}{3}$ to get $x = 9 \times \frac{4}{3} = 12$ . This isolates $x$ . Always multiply both sides by the reciprocal to maintain equality, then simplify the right side. Check your solution by substituting it back into the original equation to verify correctness.

To solve any linear equation, follow these steps: First, simplify both sides by distributing and combining like terms. For example, distribute multiplication over parentheses and add or subtract constants. Second, use addition or subtraction properties of equality to collect all variable terms on one side and constants on the other. This involves adding or subtracting terms on both sides to maintain balance. Third, use multiplication or division properties of equality to isolate the variable completely by canceling coefficients. Finally, check your solution by substituting the value back into the original equation to ensure it makes the equation true. This systematic approach helps solve linear equations efficiently and accurately.

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Solving Linear Equations: Videos & Practice Problems

Introduction to Linear Equations

Introduction to Linear Equations Video Summary

Introduction to Linear Equations Example 1

Introduction to Linear Equations Example 1 Video Summary

Identify the following as either an expression or equation.2m3+8\frac{2m}{3}+8

Identify the following as either an expression or equation.4(a−2)=214\left(a-2\right)=21

Which of the following is a linear equation in one variable?

Which of the following is a linear equation in one variable?

Verify that the given value is a solution to the equation.y=−2;5y+4=14y=-2;5y+4=14

Verify that the given value is a solution to the equation.a=2;4a+3=a+9a=2;4a+3=a+9

Addition and Subtraction Properties of Equality

Addition and Subtraction Properties of Equality Video Summary

Addition and Subtraction Properties of Equality Example 2

Addition and Subtraction Properties of Equality Example 2 Video Summary

Solve the given linear equation using addition and subtraction properties of equality.x+28=−38x+\frac28=-\frac38

Solve the given linear equation using addition and subtraction properties of equality.−5.4+c=1.6-5.4+c=1.6

Solve the given linear equation using addition and subtraction properties of equality.6h−(−12)=5+5h6h-\left(-12\right)=5+5h

Solve the given linear equation using addition and subtraction properties of equality.2(x+5)=3(x−1)2\left(x+5\right)=3\left(x-1\right)

Solve the given linear equation using addition and subtraction properties of equality.3(y+3)+(1−y)=3y+133\left(y+3\right)+\left(1-y\right)=3y+13

Addition and Subtraction Properties of Equality Example 3

Addition and Subtraction Properties of Equality Example 3 Video Summary

Multiplication and Division Properties of Equality

Multiplication and Division Properties of Equality Video Summary

Multiplication and Division Properties of Equality Example 1

Multiplication and Division Properties of Equality Example 1 Video Summary

Solve the given linear equation using multiplication and division properties of equality.−8x=64-8x=64

Solve the given linear equation using multiplication and division properties of equality.y4=−216\frac{y}{4}=-\frac{21}{6}

Solve the given linear equation using multiplication and division properties of equality.−12614=3y-\frac{126}{14}=3y

Solve the given linear equation, then check your answer−(h+3)=11-\left(h+3\right)=11

Solve the given linear equation, then check your answer0.5t+1.5t=7+3−40.5t+1.5t=7+3-4

Multiplication and Division Properties of Equality Example 2

Multiplication and Division Properties of Equality Example 2 Video Summary

Multiplication Property for Fraction Coefficients

Multiplication Property for Fraction Coefficients Video Summary

Solve the given linear equation, then check your answer.12=34x\frac12=\frac34x

Strategy for Solving Linear Equations

Strategy for Solving Linear Equations Video Summary

Solve the given linear equation. Check your solution.2(x+3)=142\left(x+3\right)=14

Solve the given linear equation. Check your solution.−5−y=3(y+9)-5-y=3\left(y+9\right)

Solve the given linear equation. Check your solution.4(x+1)−3(x−2)=2x+54\left(x+1\right)-3\left(x-2\right)=2x+5

Here’s what students ask on this topic:

What is a linear equation and how do you identify it?

What is a linear equation and how do you identify it?

How do you solve a linear equation using the addition and subtraction properties of equality?

How do you solve a linear equation using the addition and subtraction properties of equality?

What are the multiplication and division properties of equality and how are they used to solve linear equations?

What are the multiplication and division properties of equality and how are they used to solve linear equations?

How do you solve linear equations when the variable has a fractional coefficient?

How do you solve linear equations when the variable has a fractional coefficient?

What is the step-by-step process to solve any linear equation?

What is the step-by-step process to solve any linear equation?

Your Intermediate Algebra tutors

Identify the following as either an expression or equation.
$\frac{2 m}{3} + 8$

Identify the following as either an expression or equation.
$4 (a - 2) = 21$

Verify that the given value is a solution to the equation.
$y = - 2; 5 y + 4 = 14$

Verify that the given value is a solution to the equation.
$a = 2; 4 a + 3 = a + 9$

Solve the given linear equation using addition and subtraction properties of equality.
$x + \frac{2}{8} = - \frac{3}{8}$

Solve the given linear equation using addition and subtraction properties of equality.
$- 5.4 + c = 1.6$

Solve the given linear equation using addition and subtraction properties of equality.
$6 h - (- 12) = 5 + 5 h$

Solve the given linear equation using addition and subtraction properties of equality.
$2 (x + 5) = 3 (x - 1)$

Solve the given linear equation using addition and subtraction properties of equality.
$3 (y + 3) + (1 - y) = 3 y + 13$

Solve the given linear equation using multiplication and division properties of equality.
$- 8 x = 64$

Solve the given linear equation using multiplication and division properties of equality.
$\frac{y}{4} = - \frac{21}{6}$

Solve the given linear equation using multiplication and division properties of equality.
$- \frac{126}{14} = 3 y$

Solve the given linear equation, then check your answer
$- (h + 3) = 11$

Solve the given linear equation, then check your answer
$0.5 t + 1.5 t = 7 + 3 - 4$

Solve the given linear equation, then check your answer.
$\frac{1}{2} = \frac{3}{4} x$

Solve the given linear equation. Check your solution.
$2 (x + 3) = 14$

Solve the given linear equation. Check your solution.
$- 5 - y = 3 (y + 9)$

Solve the given linear equation. Check your solution.
$4 (x + 1) - 3 (x - 2) = 2 x + 5$