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Linear Equations in One Variable

Solve equations in one variable algebraically.

Linear Equations in One Variable

Simplify expressions using order of operations (IA 1.1.3)
Solve linear equations using a general strategy (IA 2.1.1)

Example

Try it.

Simplify: \(5+{2}^{3}+3[6-3(4-2)].\)

Solution


Are there any parentheses (or other grouping symbols)? Yes.
Focus on the parentheses that are inside the brackets. Subtract.
Continue inside the brackets and multiply.
Continue inside the brackets and subtract.
The expression inside the brackets requires no further simplification.
Are there any exponents? Yes. Simplify exponents.
Is there any multiplication or division? Yes.
Multiply.
Is there any addition of subtraction? Yes.
Add.
Add.

Try it.

\(3(1+9∙6)-{4}^{2}\)

Try it.

\({2}^{3}-12\div (9-5)\)

Try it.

\(33\div 3+4(7-2)\)

Try it.

\(10+3[6-2(4-2)]-{2}^{4}\)

Evaluate the following expressions being sure to follow the order of operations:

Try it.

When \(x=3\) ,

ⓐ \({x}^{5}\)
ⓑ \({5}^{x}\)
ⓒ \(3{x}^{2}-4x-8\)

Try it.

When \(x=3,y=-2\)
\(6{x}^{2}+3xy-9{y}^{2}\)

Try it.

When \(x=-8,y=3\)
\((x+y{)}^{2}\)

Simplify by combining like terms:

Try it.

\(10a+7+5a-2+7a-4\)

Try it.

\(5b+9b+10(2b+3b)+5\)

Condensed — the full section is in OpenStax College Algebra 2e.

Solving Linear Equations in One Variable

A linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form \(ax+b=0\) and are solved using basic algebraic operations.

We begin by classifying linear equations in one variable as one of three types: identity, conditional, or inconsistent. An identity equation is true for all values of the variable. Here is an example of an identity equation.

\[3x=2x+x\]

The solution set consists of all values that make the equation true. For this equation, the solution set is all real numbers because any real number substituted for \(x\) will make the equation true.

A conditional equation is true for only some values of the variable. For example, if we are to solve the equation \(5x+2=3x-6,\) we have the following:

\[\begin{array}{lll}5x+2 & = & 3x-6 \\ 2x & = & -8 \\ x & = & -4\end{array}\]

The solution set consists of one number: \(\{-4\}.\) It is the only solution and, therefore, we have solved a conditional equation.

An inconsistent equation results in a false statement. For example, if we are to solve \(5x-15=5(x-4),\) we have the following:

\[\begin{array}{llll}5x-15 & = & 5x-20 & \\ 5x-15-5x & = & 5x-20-5x & \ \text{Subtract }5x\ \text{from both sides}. \\ -15 & \ne & -20 & \ \text{False statement}\end{array}\]

Indeed, \(-15\ne \ -20.\) There is no solution because this is an inconsistent equation.

Condensed — the full section is in OpenStax College Algebra 2e.

Solving a Rational Equation

In this section, we look at rational equations that, after some manipulation, result in a linear equation. If an equation contains at least one rational expression, it is a considered a rational equation.

Recall that a rational number is the ratio of two numbers, such as \(\frac{2}{3}\) or \(\frac{7}{2}.\) A rational expression is the ratio, or quotient, of two polynomials. Here are three examples.

\[\frac{x+1}{{x}^{2}-4},\ \frac{1}{x-3},\ \text{or}\ \frac{4}{{x}^{2}+x-2}\]

Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD).

Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out.

Example

Try it.

Solve the rational equation: \(\frac{7}{2x}-\frac{5}{3x}=\frac{22}{3}.\)

Solution

We have three denominators; \(2x,3x,\) and 3. The LCD must contain \(2x,3x,\) and 3. An LCD of \(6x\) contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD \(6x.\)

\[\begin{array}{llll}(6x)(\frac{7}{2x}-\frac{5}{3x}) & = & (\frac{22}{3})(6x) & \\ (6x)(\frac{7}{2x})-(6x)(\frac{5}{3x}) & = & (\frac{22}{3})(6x) & \ \text{Use the distributive property}. \\ (6x)(\frac{7}{2x})-(6x)(\frac{5}{3x}) & = & (\frac{22}{3})(6x) & \ \text{Cancel out the common factors}. \\ 3(7)-2(5) & = & 22(2x) & \ \text{Multiply remaining factors by each numerator}. \\ 21-10 & = & 44x & \\ 11 & = & 44x & \\ \frac{11}{44} & = & x & \\ \frac{1}{4} & = & x & \end{array}\]

\[x(x-1)3=3x(x-1)\]

So, both sides of the equation would be multiplied by \(3x(x-1).\) Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out.

Another example is a problem with two denominators, such as \(x\) and \({x}^{2}+2x.\) Once the second denominator is factored as \({x}^{2}+2x=x(x+2),\) there is a common factor of x in both denominators and the LCD is \(x(x+2).\)

\[\frac{a}{b}=\frac{c}{d}\]

Condensed — the full section is in OpenStax College Algebra 2e.

Finding a Linear Equation

Perhaps the most familiar form of a linear equation is the slope-intercept form, written as \(y=mx+b,\) where \(m=\text{slope}\) and \(b=y\text{-intercept}\text{.}\) Let us begin with the slope.

The slope of a line refers to the ratio of the vertical change in y over the horizontal change in x between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run.

\[m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\]

If the slope is positive, the line slants upward to the right. If the slope is negative, the line slants downward to the right. As the absolute value of the slope increases, the line becomes steeper. Some examples are shown in . The lines indicate the following slopes: \(m=-3,\) \(m=2,\) and \(m=\frac{1}{3}.\)

Example

Try it.

Find the slope of a line that passes through the points \((2,-1)\) and \((-5,3).\)

Solution

We substitute the y-values and the x-values into the formula.

\[\begin{array}{lll}m & = & \frac{3-(-1)}{-5-2} \\ & = & \frac{4}{-7} \\ & = & -\frac{4}{7}\end{array}\]

The slope is \(-\frac{4}{7}.\)

Example

Try it.

Identify the slope and y-intercept, given the equation \(y=-\frac{3}{4}x-4.\)

Solution

As the line is in \(y=mx+b\) form, the given line has a slope of \(m=-\frac{3}{4}.\) The y-intercept is \(b=-4.\)

Condensed — the full section is in OpenStax College Algebra 2e.

Determining Whether Graphs of Lines are Parallel or Perpendicular

Parallel lines have the same slope and different y-intercepts. Lines that are parallel to each other will never intersect. For example, shows the graphs of various lines with the same slope, \(m=2.\)

All of the lines shown in the graph are parallel because they have the same slope and different y-intercepts.

Lines that are perpendicular intersect to form a \(90^{\circ}\) -angle. The slope of one line is the negative reciprocal of the other. We can show that two lines are perpendicular if the product of the two slopes is \(-1:{m}_{1}⋅{m}_{2}=-1.\) For example, shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of \(-\frac{1}{3}.\)

\[\begin{array}{lll}{m}_{1}⋅{m}_{2} & = & -1 \\ 3⋅(-\frac{1}{3}) & = & -1\end{array}\]

Example

Try it.

Graph the equations of the given lines, and state whether they are parallel, perpendicular, or neither: \(3y=-4x+3\) and \(3x-4y=8.\)

Solution

The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form.

First equation:

\[\begin{array}{lll}3y & = & -4x+3 \\ y & = & -\frac{4}{3}x+1\end{array}\]

Second equation:

\[\begin{array}{lll}3x-4y & = & 8 \\ -4y & = & -3x+8 \\ y & = & \frac{3}{4}x-2\end{array}\]

See the graph of both lines in

From the graph, we can see that the lines appear perpendicular, but we must compare the slopes.

\[\begin{array}{lll}{m}_{1} & = & -\frac{4}{3} \\ {m}_{2} & = & \frac{3}{4} \\ {m}_{1}⋅{m}_{2} & = & (-\frac{4}{3})(\frac{3}{4})=-1\end{array}\]

The slopes are negative reciprocals of each other, confirming that the lines are perpendicular.

Writing the Equations of Lines Parallel or Perpendicular to a Given Line

As we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use the point-slope formula to write the equation of the new line.

Example

Try it.

Write the equation of line parallel to a \(5x+3y=1\) and passing through the point \((3,5).\)

Solution

First, we will write the equation in slope-intercept form to find the slope.

\[\begin{array}{lll}5x+3y & = & 1 \\ 3y & = & -5x+1 \\ y & = & -\frac{5}{3}x+\frac{1}{3}\end{array}\]

The slope is \(m=-\frac{5}{3}.\) The y-intercept is \(\frac{1}{3},\) but that really does not enter into our problem, as the only thing we need for two lines to be parallel is the same slope. The one exception is that if the y-intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point with the point-slope formula.

\[\begin{array}{lll}y-5 & = & -\frac{5}{3}(x-3) \\ y-5 & = & -\frac{5}{3}x+5 \\ y & = & -\frac{5}{3}x+10\end{array}\]

The equation of the line is \(y=-\frac{5}{3}x+10.\) See .

Example

Try it.

Find the equation of the line perpendicular to \(5x-3y+4=0\) and passing through the point \((-4,1).\)

Solution

The first step is to write the equation in slope-intercept form.

\[\begin{array}{lll}5x-3y+4 & = & 0 \\ -3y & = & -5x-4 \\ y & = & \frac{5}{3}x+\frac{4}{3}\end{array}\]

We see that the slope is \(m=\frac{5}{3}.\) This means that the slope of the line perpendicular to the given line is the negative reciprocal, or \(-\frac{3}{5}.\) Next, we use the point-slope formula with this new slope and the given point.

\[\begin{array}{lll}y-1 & = & -\frac{3}{5}(x-(-4)) \\ y-1 & = & -\frac{3}{5}x-\frac{12}{5} \\ y & = & -\frac{3}{5}x-\frac{12}{5}+\frac{5}{5} \\ y & = & -\frac{3}{5}x-\frac{7}{5}\end{array}\]

Condensed — the full section is in OpenStax College Algebra 2e.

Key Concepts

We can solve linear equations in one variable in the form \(ax+b=0\) using standard algebraic properties. See and .
A rational expression is a quotient of two polynomials. We use the LCD to clear the fractions from an equation. See and .
All solutions to a rational equation should be verified within the original equation to avoid an undefined term, or zero in the denominator. See and and .
Given two points, we can find the slope of a line using the slope formula. See .
We can identify the slope and y-intercept of an equation in slope-intercept form. See .
We can find the equation of a line given the slope and a point. See .
We can also find the equation of a line given two points. Find the slope and use the point-slope formula. See .
The standard form of a line has no fractions. See .
Horizontal lines have a slope of zero and are defined as \(y=c,\) where c is a constant.
Vertical lines have an undefined slope (zero in the denominator), and are defined as \(x=c,\) where c is a constant. See .
Parallel lines have the same slope and different y-intercepts. See and .
Perpendicular lines have slopes that are negative reciprocals of each other unless one is horizontal and the other is vertical. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Simplify: \(5+{2}^{3}+3[6-3(4-2)].\)

ჲრკპთირვ ჲრდჲგჲპა.


Are there any parentheses (or other grouping symbols)? Yes.
Focus on the parentheses that are inside the brackets. Subtract.
Continue inside the brackets and multiply.
Continue inside the brackets and subtract.
The expression inside the brackets requires no further simplification.
Are there any exponents? Yes. Simplify exponents.
Is there any multiplication or division? Yes.
Multiply.
Is there any addition of subtraction? Yes.
Add.
Add.

\(3(1+9∙6)-{4}^{2}\)
\({2}^{3}-12\div (9-5)\)
\(33\div 3+4(7-2)\)
\(10+3[6-2(4-2)]-{2}^{4}\)
When \(x=3,y=-2\)
\(6{x}^{2}+3xy-9{y}^{2}\)
When \(x=-8,y=3\)
\((x+y{)}^{2}\)
\(10a+7+5a-2+7a-4\)
\(5b+9b+10(2b+3b)+5\)

\(2(w+5)+1=10+4w+2\)

ჲრკპთირვ ჲრდჲგჲპა.

Use distributive property to remove parentheses:	\(2w+10+1=10+4w+2\)
Combine like terms on each side:	\(2w+11=12+4w\)
Subtract \(2w\) from each side to bring variables to one side:	\(2w-2w+11=12+4w-2w\)
Combine like terms:	\(11=12+2w\)
Subtract 12 from each side to bring constants to one side:	\(11-12=12-12+2w\)
Combine like terms:	\(-1=2w\)
Divide each side by 2 to isolate the variable terms:	\(\frac{-1}{2}=\frac{2w}{2}\)
Simplify:	\(\frac{-1}{2}=w\) or \(w=-\frac{1}{2}\)
To check your solution, replace \(w\) with \(-\frac{1}{2}\) in the original equation and simplify:	\(2(w+5)+1=10+4w+2\)
	\(2(-\frac{1}{2}+5)+1=10+4(-\frac{1}{2})+2\)
	\(2(\frac{9}{2})+1=10+(-2)+2\) \(9+1=8+2\)
The solution checks, we reached a true statement.	\(10=10\)

\(15(y-9)=-60\)
\(-2(11-7x)+54=4\)
\(3(4n-1)-2=8n+3\)
\(12+2(5-3y)=-9(y-1)-2\)
\(\frac{1}{4}(20x+12)=x+7\)
\(22(3m-4)=8(2m+9)\)
\(\frac{3x+4}{2}+1=\frac{5x+10}{8}\)
\(0.05n+0.10(n+8)=2.15\)
Solve the following equation: \(2x+7=19.\)

ჲრკპთირვ ჲრდჲგჲპა.

This equation can be written in the form \(ax+b=0\) by subtracting \(19\) from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations.
\[\begin{array}{llll}2x+7 & = & 19 & \\ 2x & = & 12 & \ \text{Subtract 7 from both sides}\text{.} \\ x & = & 6 & \ \text{Multiply both sides by }\frac{1}{2}\ \text{or divide by 2}\text{.}\end{array}\]
The solution is 6.
Solve the linear equation in one variable: \(2x+1=-9.\)

ჲრკპთირვ ჲრდჲგჲპა.

\(x=-5\)
Solve the following equation: \(4(x-3)+12=15-5(x+6).\)

ჲრკპთირვ ჲრდჲგჲპა.

Apply standard algebraic properties.
\[\begin{array}{llll}4(x-3)+12 & = & 15-5(x+6) & \\ 4x-12+12 & = & 15-5x-30 & \ \text{Apply the distributive property}\text{.} \\ 4x & = & -15-5x & \ \text{Combine like terms}. \\ 9x & = & -15 & \ \text{Place }x\text{-terms on one side and simplify}. \\ x & = & -\frac{15}{9} & \ \text{Multiply both sides by }\frac{1}{9}\text{, the reciprocal of 9}. \\ x & = & -\frac{5}{3} & \end{array}\]
Solve the equation in one variable: \(-2(3x-1)+x=14-x.\)

ჲრკპთირვ ჲრდჲგჲპა.

\(x=-3\)
Solve the rational equation: \(\frac{7}{2x}-\frac{5}{3x}=\frac{22}{3}.\)

ჲრკპთირვ ჲრდჲგჲპა.

We have three denominators; \(2x,3x,\) and 3. The LCD must contain \(2x,3x,\) and 3. An LCD of \(6x\) contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD \(6x.\)
\[\begin{array}{llll}(6x)(\frac{7}{2x}-\frac{5}{3x}) & = & (\frac{22}{3})(6x) & \\ (6x)(\frac{7}{2x})-(6x)(\frac{5}{3x}) & = & (\frac{22}{3})(6x) & \ \text{Use the distributive property}. \\ (6x)(\frac{7}{2x})-(6x)(\frac{5}{3x}) & = & (\frac{22}{3})(6x) & \ \text{Cancel out the common factors}. \\ 3(7)-2(5) & = & 22(2x) & \ \text{Multiply remaining factors by each numerator}. \\ 21-10 & = & 44x & \\ 11 & = & 44x & \\ \frac{11}{44} & = & x & \\ \frac{1}{4} & = & x & \end{array}\]
Solve the following rational equation:
\[\frac{2}{x}-\frac{3}{2}=\frac{7}{2x}\]

ჲრკპთირვ ჲრდჲგჲპა.

We have three denominators: \(x,\) \(2,\) and \(2x.\) No factoring is required. The product of the first two denominators is equal to the third denominator, so, the LCD is \(2x.\) Only one value is excluded from a solution set, 0. \(\) Next, multiply the whole equation (both sides of the equal sign) by \(2x.\)
\[\begin{array}{llll}2x(\frac{2}{x}-\frac{3}{2}) & = & (\frac{7}{2x})2x & \\ 2x(\frac{2}{x})-2x(\frac{3}{2}) & = & (\frac{7}{2x})2x & \ \text{Distribute }2x\text{.} \\ 2(2)-3x & = & 7 & \ \text{Denominators cancel out}. \\ 4-3x & = & 7 & \\ -3x & = & 3 & \\ x & = & -1 & \\ & & \text{or}\ \{-1\} & \end{array}\]
The proposed solution is −1, \(\) which is not an excluded value, so the solution set contains one number, \(-1,\) or \(\{-1\}\) written in set notation.
Solve the rational equation: \(\frac{2}{3x}=\frac{1}{4}-\frac{1}{6x}.\)

ჲრკპთირვ ჲრდჲგჲპა.

\(x=\frac{10}{3}\)
Solve the following rational equation: \(\frac{1}{x}=\frac{1}{10}-\frac{3}{4x}.\)

ჲრკპთირვ ჲრდჲგჲპა.

First find the common denominator. The three denominators in factored form are \(x,10=2⋅5,\) and \(4x=2⋅2⋅x.\) The smallest expression that is divisible by each one of the denominators is \(20x.\) Only \(x=0\) is an excluded value. Multiply the whole equation by \(20x.\)
\[\begin{array}{lll}20x(\frac{1}{x}) & = & (\frac{1}{10}-\frac{3}{4x})20x \\ 20 & = & 2x-15 \\ 35 & = & 2x \\ \frac{35}{2} & = & x\end{array}\]
The solution is \(\frac{35}{2}.\)
Solve the rational equation: \(-\frac{5}{2x}+\frac{3}{4x}=-\frac{7}{4}.\)

ჲრკპთირვ ჲრდჲგჲპა.

\(x=1\)
Solve the following rational equations and state the excluded values:
1. ⓐ \(\frac{3}{x-6}=\frac{5}{x}\)
2. ⓑ \(\frac{x}{x-3}=\frac{5}{x-3}-\frac{1}{2}\)
3. ⓒ \(\frac{x}{x-2}=\frac{5}{x-2}-\frac{1}{2}\)
ჲრკპთირვ ჲრდჲგჲპა.
1. ⓐ
  The denominators \(x\) and \(x-6\) have nothing in common. Therefore, the LCD is the product \(x(x-6).\) However, for this problem, we can cross-multiply.
  \[\begin{array}{llll}\frac{3}{x-6} & = & \frac{5}{x} & \\ 3x & = & 5(x-6) & \ \text{Distribute}\text{.} \\ 3x & = & 5x-30 & \\ -2x & = & -30 & \\ x & = & 15 & \end{array}\]
  The solution is 15. \(\) The excluded values are \(6\) and \(0.\)
2. ⓑ
  The LCD is \(2(x-3).\) Multiply both sides of the equation by \(2(x-3).\)
  \[\begin{array}{lll}2(x-3)(\frac{x}{x-3}) & = & (\frac{5}{x-3}-\frac{1}{2})2(x-3) \\ \frac{2(x-3)x}{x-3} & = & \frac{2(x-3)5}{x-3}-\frac{2(x-3)}{2} \\ 2x & = & 10-(x-3) \\ 2x & = & 10-x+3 \\ 2x & = & 13-x \\ 3x & = & 13 \\ x & = & \frac{13}{3}\end{array}\]
  The solution is \(\frac{13}{3}.\) The excluded value is \(3.\)
3. ⓒ
  The least common denominator is \(2(x-2).\) Multiply both sides of the equation by \(2(x-2).\)
  \[\begin{array}{lll}2(x-2)(\frac{x}{x-2}) & = & (\frac{5}{x-2}-\frac{1}{2})2(x-2) \\ 2x & = & 10-(x-2) \\ 2x & = & 12-x \\ 3x & = & 12 \\ x & = & 4\end{array}\]
  The solution is 4. The excluded value is \(2.\)
Solve \(\frac{-3}{2x+1}=\frac{4}{3x+1}.\) State the excluded values.

ჲრკპთირვ ჲრდჲგჲპა.

\(x=-\frac{7}{17}.\) Excluded values are \(x=-\frac{1}{2}\) and \(x=-\frac{1}{3}.\)
Solve the rational equation after factoring the denominators: \(\frac{2}{x+1}-\frac{1}{x-1}=\frac{2x}{{x}^{2}-1}.\) State the excluded values.

ჲრკპთირვ ჲრდჲგჲპა.

We must factor the denominator \({x}^{2}-1.\) We recognize this as the difference of squares, and factor it as \((x-1)(x+1).\) Thus, the LCD that contains each denominator is \((x-1)(x+1).\) Multiply the whole equation by the LCD, cancel out the denominators, and solve the remaining equation.
\[\begin{array}{lll}(x-1)(x+1)(\frac{2}{x+1}-\frac{1}{x-1}) & = & (\frac{2x}{(x-1)(x+1)})(x-1)(x+1) \\ 2(x-1)-1(x+1) & = & 2x \\ 2x-2-x-1 & = & 2x\ \text{Distribute the negative sign}. \\ -3-x & = & 0 \\ -3 & = & x\end{array}\]
The solution is \(-3.\) The excluded values are \(1\) and \(-1.\)
Solve the rational equation: \(\frac{2}{x-2}+\frac{1}{x+1}=\frac{1}{{x}^{2}-x-2}.\)

ჲრკპთირვ ჲრდჲგჲპა.

\(x=\frac{1}{3}\)
Find the slope of a line that passes through the points \((2,-1)\) and \((-5,3).\)

ჲრკპთირვ ჲრდჲგჲპა.

We substitute the y-values and the x-values into the formula.
\[\begin{array}{lll}m & = & \frac{3-(-1)}{-5-2} \\ & = & \frac{4}{-7} \\ & = & -\frac{4}{7}\end{array}\]
The slope is \(-\frac{4}{7}.\)
Find the slope of the line that passes through the points \((-2,6)\) and \((1,4).\)

ჲრკპთირვ ჲრდჲგჲპა.

\(m=-\frac{2}{3}\)
Identify the slope and y-intercept, given the equation \(y=-\frac{3}{4}x-4.\)

ჲრკპთირვ ჲრდჲგჲპა.

As the line is in \(y=mx+b\) form, the given line has a slope of \(m=-\frac{3}{4}.\) The y-intercept is \(b=-4.\)
Write the equation of the line with slope \(m=-3\) and passing through the point \((4,8).\) Write the final equation in slope-intercept form.

ჲრკპთირვ ჲრდჲგჲპა.

Using the point-slope formula, substitute \(-3\) for m and the point \((4,8)\) for \(({x}_{1},{y}_{1}).\)
\[\begin{array}{lll}y-{y}_{1} & = & m(x-{x}_{1}) \\ y-8 & = & -3(x-4) \\ y-8 & = & -3x+12 \\ y & = & -3x+20\end{array}\]
Given \(m=4,\) find the equation of the line in slope-intercept form passing through the point \((2,5).\)

ჲრკპთირვ ჲრდჲგჲპა.

\(y=4x-3\)
Find the equation of the line passing through the points \((3,4)\) and \((0,-3).\) Write the final equation in slope-intercept form.

ჲრკპთირვ ჲრდჲგჲპა.

First, we calculate the slope using the slope formula and two points.
\[\begin{array}{lll}m & = & \frac{-3-4}{0-3} \\ & = & \frac{-7}{-3} \\ & = & \frac{7}{3}\end{array}\]
Next, we use the point-slope formula with the slope of \(\frac{7}{3},\) and either point. Let’s pick the point \((3,4)\) for \(({x}_{1},{y}_{1}).\)
\[\begin{array}{lll}y-4 & = & \frac{7}{3}(x-3) \\ y-4 & = & \frac{7}{3}x-7\ \text{Distribute the }\frac{7}{3}. \\ y & = & \frac{7}{3}x-3\end{array}\]
In slope-intercept form, the equation is written as \(y=\frac{7}{3}x-3.\)
Find the equation of the line with \(m=-6\) and passing through the point \((\frac{1}{4},-2).\) Write the equation in standard form.

ჲრკპთირვ ჲრდჲგჲპა.

We begin using the point-slope formula.
\[\begin{array}{lll}y-(-2) & = & -6(x-\frac{1}{4}) \\ y+2 & = & -6x+\frac{3}{2}\end{array}\]
From here, we multiply through by 2, as no fractions are permitted in standard form, and then move both variables to the left aside of the equal sign and move the constants to the right.
\[\begin{array}{lll}2(y+2) & = & (-6x+\frac{3}{2})2 \\ 2y+4 & = & -12x+3 \\ 12x+2y & = & -1\end{array}\]
This equation is now written in standard form.
Find the equation of the line in standard form with slope \(m=-\frac{1}{3}\) and passing through the point \((1,\frac{1}{3}).\)

ჲრკპთირვ ჲრდჲგჲპა.

\(x+3y=2\)

Symbols used here

^\circ

degrees
1/360 of a full turn. 180° = π radians.

\neq

not equal
The two sides are different.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Linear Equations in One Variable

Solve equations in one variable algebraically.
Solve a rational equation.
Find a linear equation.
Given the equations of two lines, determine whether their graphs are parallel or perpendicular.
Write the equation of a line parallel or perpendicular to a given line.
Simplify expressions using order of operations (IA 1.1.3)
Solve linear equations using a general strategy (IA 2.1.1)
Parentheses and Other Grouping Symbols

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

ჲოთრაი ჟამ.

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

მეტი Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value