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Solving Linear Equations

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Solving linear equations means to find the solution of a linear equation. Here, the methods of solving linear equations are explained for its three main types which include linear equations in one variable, linear equations in two variables and linear equations in three variables.

What Does Solving Linear Equations Mean?

Solving a linear equation refers to finding the solution of linear equations in one, two, three or variables. In simple words, a solution of a linear equation means the value or values of the variables involved in the equation.

How to Solve Linear Equations?

There are six main methods to solve linear equations. These methods for finding the solution of linear equations are:

Graphical Method of Solving Linear Equations

To solve linear equations graphically, first graph both equations in the same coordinate system and check for the intersection point in the graph. For example, take two equations as 2x + 3y = 9 and x – y = 3.

Now, to plot the graph, consider x = {0. 1, 2, 3, 4} and solve for y. Once (x, y) is obtained, plot the points on the graph. It should be noted that by having more values of x and y will make the graph more accurate.

Check: Graphical Method of Solving Linear Programming

The graph of 2x + 3y = 9 and x – y = 3 will be as follows:

Graphical Method of Solving Linear Equations

In the graph, check for the intersection point of both the lines. Here, it is mentioned as (x, y). Check the value of that point and that will be the solution of both the given equations . Here, the value of (x, y) = (3.6, 0.6).

Elimination Method of Solving Linear Equations

In the elimination method , any of the coefficients is first equated and eliminated. After elimination, the equations are solved to obtain the other equation. Below is an example of solving linear equations using the elimination method for better understanding.

Consider the same equations as

2x + 3y = 9 ———–(i)

x – y = 3 ———–(ii)

Here, if equation (ii) is multiplied by 2, the coefficient of “x” will become the same and can be subtracted.

So, multiply equation (ii) × 2 and then subtract equation (i)

Or, y = ⅗ = 0.6

Now, put the value of y = 0.6 in equation (ii).

So, x – 0.6 = 3

In this way, the value of x, y is found to be 3.6 and 0.6.

Substitution Method of Solving Linear Equations

To solve a linear equation using the substitution method , first, isolate the value of one variable from any of the equations. Then, substitute the value of the isolated variable in the second equation and solve it. Take the same equations again for example.

Now, consider equation (ii) and isolate the variable “x”.

So, equation (ii) becomes,

Now, substitute the value of x in equation (i). So, equation (i) will be-

⇒ 2(3 + y) + 3y = 9

⇒ 6 + 2y + 3y = 9

Now, substitute “y” value in equation (ii).

⇒ x = 3 + 0.6

Thus, (x, y) = (3.6, 0.6).

Cross Multiplication Method of Solving Linear Equations

Linear equations can be easily solved using the cross multiplication method. In this method, the cross-multiplication technique is used to simplify the solution. For the cross-multiplication method for solving 2 variable equation, the formula used is:

x /(b 1 c 2 − b 2 c 1 ) = y / (c 1 a 2 − c 2 a 1 ) = 1 /(b 2 a 1 − b 1 a 2 )

For example, consider the equations

a 1 = 2, b 1 = 3, c 1 = -9

a 2 = 1, b 2 = -1, c 2 = -3

Now, solve using the aforementioned formula.

x = (b 1 c 2 − b 2 c 1 ) / (b 2 a 1 − b 1 a 2 )

Putting the respective value we get,

Similarly, solve for y.

y = (c 1 a 2 − c 2 a 1 ) / (b 2 a 1 − b 1 a 2 )

So, y = ⅗ = 0.6

Matrix Method of Solving Linear Equations

Linear equations can also be solved using matrix method. This method is extremely helpful for solving linear equations in two or three variables. Consider three equations as:

a 1 x + a 2 y + a 3 z = d 1

b 1 x + b 2 y + b 3 z = d 2

c 1 x + c 2 y + c 3 z = d 3

These equations can be written as:

Matrix Method of Solving Linear Equations

⇒ AX = B ————-(i)

Here, the A matrix, B matrix and X matrix are:

Solving Linear Equations Using Matrix Method

Now, multiply (i) by A -1 to get:

A −1 AX = A −1 B ⇒ I.X = A −1 B

⇒ X = A −1 B

Learn more on how to solve linear equations with matrix method here .

Determinant Method of Solving Linear Equations (Cramer’s Rule)

Determinants method can be used to solve linear equations in two or three variables easily. For two variables and three variables of linear equations, the procedure is as follows.

For Linear Equations in Two Variables:

Or, x = (b 1 c 2 − b 2 c 1 ) / (b 2 a 1 − b 1 a 2 ) and y = (c 1 a 2 − c 2 a 1 ) / (b 2 a 1 − b 1 a 2 )

Determinant Method of Solving Linear Equations in Two Variables

For Linear Equations in Three Variables:

Determinant Method of Solving Linear Equations in Three Variables

Learn more on how to solve linear equations with determinants method here .

Related Video: Solving an Equation

linear equations solve method

Methods of Solving Linear Equations in One Variable

Solving a linear equation with one variable is extremely easy and quick. To solve any two equations having only 1 variable, bring all the variable terms on one side and the constants on the other. The graphical method can also be used in which the point of intersection of the line with the x-axis or y-axis will give the solution of the equation.

For example, consider the equation 2x + 4 + 7 = 4x – 3 + x

Here, combine the “x” terms and bring them on one side.

Methods of Solving Linear Equations in Two Variables

To solve a linear equation in two variables , any of the above-mentioned methods can be used i.e. graphical method, elimination method, substitution method, cross multiplication method, matrix method, determinants method.

Methods of Solving Linear Equations in Three or More Variables

For solving any equation having three or more variables, the graphical, elimination and the substitution method is not feasible. For solving a three-variable equation, the cross-multiplication method is the most preferred method. Even matrix Cramer’s rule is extremely useful for solving equations having 3 or more variables.

Check: Solve Linear Equation in Two Or Three Variables

Topics Related to Solution of Linear Equations:

Frequently asked questions, what is a linear equation.

A linear equation is an equation where each variable has a degree of one. An example of a linear equation is 4x + 3y = 10.

What are the Methods of Solving Linear Equations?

The 6 most common methods of solving a linear equation are:

How to Solve Linear Equations with Fractions?

To solve a linear equation with fraction, follow these steps:

Learn more about fraction here .

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

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Plotting the ordered pair solutions and drawing the line:

Practice Problem 1a: Solve the system by graphing.

Practice Problem 2a: Solve the system by the substitution method.

Practice Problem 3a: Solve the system by the elimination method.

Go to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions.

Solving Linear Equations

Solving linear equations means finding the value of the variable(s) given in the linear equations. A linear equation is a combination of an algebraic expression and an equal to (=) symbol. It has a degree of 1 or it can be called a first-degree equation. For example, x + y = 4 is a linear equation. Sometimes, we may have to find the values of variables involved in a linear equation. When we are given two or more such linear equations, we can find the values of each variable by solving linear equations. There are a few methods to solve linear equations. Let us discuss each of these methods in detail.

Solving Linear Equations in One Variable

A linear equation in one variable is an equation of degree one and has only one variable term. It is of the form 'ax+b = 0', where 'a' is a non zero number and 'x' is a variable. By solving linear equations in one variable, we get only one solution for the given variable. An example for this is 3x - 6 = 0. The variable 'x' has only one solution, which is calculated as 3x - 6 = 0 3x = 6 x = 6/3 x = 2

For solving linear equations with one variable, simplify the equation such that all the variable terms are brought to one side and the constant value is brought to the other side. If there are any fractional terms then find the LCM ( Least Common Multiple ) and simplify them such that the variable terms are on one side and the constant terms are on the other side. Let us work out a small example to understand this.

4x + 8 = 8x - 10. To find the value of 'x', let us simplify and bring the 'x' terms to one side and the constant terms to another side.

4x - 8x = -10 - 8 -4x = -18 4x = 18 x = 18/4 On simplifying, we get x = 9/2.

Solving Linear Equations by Substitution Method

The substitution method is one of the methods of solving linear equations. In the substitution method , we rearrange the equation such that one of the values is substituted in the second equation. Now that we are left with an equation that has only one variable, we can solve it and find the value of that variable. In the two given equations, any equation can be taken and the value of a variable can be found and substituted in another equation. For solving linear equations using the substitution method, follow the steps mentioned below. Let us understand this with an example of solving the following system of linear equations. x + y = 6 --------------(1) 2x + 4y = 20 -----------(2)

Step 1: Find the value of one of the variables using any one of the equations. In this case, let us find the value of 'x' from equation (1). x + y = 6 ---------(1) x = 6 - y Step 2: Substitute the value of the variable found in step 1 in the second linear equation. Now, let us substitute the value of 'x' in the second equation 2x + 4y = 20.

x = 6 - y Substituting the value of 'x' in 2x + 4y = 20, we get,

2(6 - y) + 4y = 20 12 - 2y + 4y = 20 12 + 2y = 20 2y = 20 - 12 2y = 8 y = 8/2 y = 4 Step 3: Now substitute the value of 'y' in either equation (1) or (2). Let us substitute the value of 'y' in equation (1).

x + y = 6 x + 4 = 6 x = 6 - 4 x = 2 Therefore, by substitution method, the linear equations are solved, and the value of x is 2 and y is 4.

Solving Linear Equations by Elimination Method

The elimination method is another way to solve a system of linear equations. Here we make an attempt to multiply either the 'x' variable term or the 'y' variable term with a constant value such that either the 'x' variable terms or the 'y' variable terms cancel out and gives us the value of the other variable. Let us understand the steps of solving linear equations by elimination method . Consider the given linear equations: 2x + y = 11 ----------- (1) x + 3y = 18 ---------- (2) Step 1: Check whether the terms are arranged in a way such that the 'x' term is followed by a 'y' term and an equal to sign and after the equal to sign the constant term should be present. The given set of linear equations are already arranged in the correct way which is ax+by=c or ax+by-c=0.

Step 2: The next step is to multiply either one or both the equations by a constant value such that it will make either the 'x' terms or the 'y' terms cancel out which would help us find the value of the other variable. Now in equation (2), let us multiply every term by the number 2 to make the coefficients of x the same in both the equations. x + 3y = 18 ---------- (2) Multiplying all the terms in equation (2) by 2, we get,

2(x) + 2(3y) = 2(18). Now equation (2) becomes, 2x + 6y = 36 -----------(2)

Elimination Method of solving linear equations

Therefore, y = 5. Step 4: Using the value obtained in step 3, find out the value of another variable by substituting the value in any of the equations. Let us substitute the value of 'y' in equation (1). We get, 2x + y = 11 2x + 5 = 11 2x = 11 - 5 2x = 6 x = 6/2 x = 3

Therefore, by solving linear equations, we get the value of x = 3 and y = 5.

Graphical Method of Solving Linear Equations

Another method for solving linear equations is by using the graph. When we are given a system of linear equations, we graph both the equations by finding values for 'y' for different values of 'x' in the coordinate system. Once it is done, we find the point of intersection of these two lines. The (x,y) values at the point of intersection give the solution for these linear equations. Let us take two linear equations and solve them using the graphical method.

x + y = 8 -------(1)

y = x + 2 --------(2)

Let us take some values for 'x' and find the values for 'y' for the equation x + y = 8. This can also be rewritten as y = 8 - x.

Let us take some values for 'x' and find the values for 'y' in the equation y = x + 2.

Plotting these points on the coordinate plane, we get a graph like this.

Graphical Method of Solving Linear Equations

Now, we find the point of intersection of these lines to find the values of 'x' and 'y'. The two lines intersect at the point (3,5). Therefore, x = 3 and y = 5 by using the graphical method of solving linear equations .

This method is also used to find the optimal solution of linear programming problems. Let us look at one more method of solving linear equations, which is the cross multiplication method.

Cross Multiplication Method of Solving Linear Equations

The cross multiplication method enables us to solve linear equations by picking the coefficients of all the terms ('x' , 'y' and the constant terms) in the format shown below and apply the formula for finding the values of 'x' and 'y'.

Cross Multiplication Method of solving linear equations

Topics Related to Solving Linear Equations

Check the given articles related to solving linear equations.

Solving Linear Equations Examples

Example 1: Solve the following linear equations by the substitution method.

3x + y = 13 --------- (1) 2x + 3y = 18 -------- (2)

By using the substitution method of solving linear equations, let us take the first equation and find the value of 'y' and substitute it in the second equation.

From equation (1), y = 13-3x. Now, substituting the value of 'y' in equation (2), we get, 2x + 3 (13 - 3x) = 18 2x + 39 - 9x = 18 -7x + 39 = 18 -7x = 18 - 39 -7x = -21 x = -21/-7 x = 3 Now, let us substitute the value of 'x = 3' in equation (1) and find the value of 'y'. 3x + y = 13 ------- (1) 3(3) + y = 13 9 + y = 13 y = 13 - 9 y = 4

Therefore, by the substitution method, the value of x is 3 and y is 4.

Example 2: Using the elimination method of solving linear equations find the values of 'x' and 'y'.

3x + y = 21 ------ (1) 2x + 3y = 28 -------- (2)

By using the elimination method, let us make the 'y' variable to be the same in both the equations (1) and (2). To do this let us multiply all the terms of the first equation by 3. Therefore equation (1) becomes,

3(3x) + 3(y) = 63 9x + 3y = 63 ---------- (3) The second equation is, 2x + 3y = 28 Now let us cancel the 'y' terms and find the value of 'x' by subtracting equation (2) from equation (3). This is done by changing the signs of all the terms in equation (2).

Solving Linear Equations Example

Example 3: Using the cross multiplication method of solving linear equations, solve the following equations.

x + 2y - 16 = 0 --------- (1) 4x - y - 10 = 0 ---------- (2)

Compare the given equation with \(a_{1}\)x + \(b_{1}\)y + \(c_{1}\) = 0, and \(a_{2}\)x+\(b_{2}\)y+\(c_{2}\) = 0. From the given equations,

\(a_{1}\) = 1, \(a_{2}\) = 4, \(b_{1}\) = 2, \(b_{2}\) = -1, \(c_{1}\) = -16, and \(c_{2}\) = -10.

By cross multiplication method,

x = \(b_{1}\)\(c_{2}\) - \(b_{2}\)\(c_{1}\)/\(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\) y = \(c_{1}\)\(a_{2}\) - \(c_{2}\)\(a_{1}\) / \(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\)

Substituting the values in the formula we get,

x = ((2)(-10)) - ((-1)(-16)) / ((1)(-1)) - ((4)(2)) x = (-20-16)/(-1-8) x = -36/-9 x = 36/9 x = 4 y = ((-16)(4)) - ((-10)(1)) / ((1)(-1)) - ((4)(2)) y = (-64 + 10) / (-1 - 8) y = -54 / -9 y = 54/9 y = 6 Therefore, by the cross multiplication method, the value of x is 4 and y is 6.

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Practice Questions on Solving Linear Equations

Faqs on solving linear equations, what does it mean by solving linear equations.

An equation that has a degree of 1 is called a linear equation. We can have one variable linear equations , two-variable linear equations , linear equations with three variables, and more depending on the number of variables in it. Solving linear equations means finding the values of all the variables present in the equation. This can be done by substitution method, elimination method, graphical method, and the cross multiplication method . All these methods are different ways of finding the values of the variables.

How to Use the Substitution Method for Solving Linear Equations?

The substitution method of solving equations states that for a given system of linear equations, find the value of either 'x' or 'y' from any of the given equations and then substitute the value found of 'x' or 'y' in another equation so that the other unknown value can be found.

How to Use the Elimination Method for Solving Linear Equations?

In the elimination method of solving linear equations, we multiply a constant or a number with one equation or both the equations such that either the 'x' terms or the 'y' terms are the same. Then we cancel out the same term in both the equations by either adding or subtracting them and find the value of one variable (either 'x' or 'y'). After finding one of the values, we substitute the value in one of the equations and find the other unknown value.

What is the Graphical Method of Solving Linear Equations?

In the graphical method of solving linear equations, we find the value of 'y' from the given equations by putting the values of x as 0, 1, 2, 3, and so on, and plot a graph in the coordinate system for the line for various values of 'x' for both the system of linear equations. We will see that these two lines intersect at a point. This point is the solution for the given system of linear equations. If there is no intersection point between two lines, then we consider them as parallel lines , and if we found that both the lines lie on each other, those are known as coincident lines and have infinitely many solutions.

What are the Steps of Solving Linear Equations that has One Variable?

A linear equation is an equation with degree 1. To solve a linear equation that has one variable we bring the variable to one side and the constant value to the other side. Then, a non-zero number may be added, subtracted, multiplied, or divided on both sides of the equation. For example, a linear equation with one variable will be of the form 'x - 4 = 2'. To find the value of 'x', we add the constant value '4' to both sides of the equation. Therefore, the value of 'x = 6'.

What are the Steps of Solving Linear Equations having Three Variables?

To solve a system of linear equations that has three variables, we take any two equations and variables. We then take another pair of linear equations and also solve for the same variable. Now that, we have two linear equations with two variables, we can use the substitution method or elimination method, or any other method to solve the values of two unknown variables. After finding these two variables, we substitute them in any of the three equations to find the third unknown variable.

What are the 4 Methods of Solving Linear Equations?

The methods for solving linear equations are given below:

Solving Systems of Linear Equations

A system of linear equations is just a set of two or more linear equations.

In two variables ( x and y ) , the graph of a system of two equations is a pair of lines in the plane.

There are three possibilities:

Zero solutions:

y = − 2 x + 4 y = − 2 x − 3

y = 0.5 x + 2 y = − 2 x − 3

Infinitely many solutions:

y = − 2 x − 4 y + 4 = − 2 x

There are a few different methods of solving systems of linear equations:

Solve the system { 3 x + 2 y = 16 7 x + y = 19

Substitute 19 − 7 x for y in the first equation and solve for x .

3 x + 2 ( 19 − 7 x ) = 16 3 x + 38 − 14 x = 16 − 11 x = − 22 x = 2

Substitute 2 for x in y = 19 − 7 x and solve for y .

y = 19 − 7 ( 2 ) y = 5

The Linear Combination Method , aka The Addition Method , aka The Elimination Method. Add (or subtract) a multiple of one equation to (or from) the other equation, in such a way that either the x -terms or the y -terms cancel out. Then solve for x (or y , whichever's left) and substitute back to get the other coordinate.

Solve the system { 4 x + 3 y = − 2 8 x − 2 y = 12

− 8 x − 6 y = 4 8 x − 2 y = 12 _ − 8 y = 16

Substitute for y in either of the original equations and solve for x .

4 x + 3 ( − 2 ) = − 2 4 x − 6 = − 2 4 x = 4 x = 1

The solution is ( 1 , − 2 ) .

The Matrix Method . This is really just the Linear Combination method, made simpler by shorthand notation.

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The elimination method for solving linear systems

Another way of solving a linear system is to use the elimination method. In the elimination method you either add or subtract the equations to get an equation in one variable.

When the coefficients of one variable are opposites you add the equations to eliminate a variable and when the coefficients of one variable are equal you subtract the equations to eliminate a variable.

\begin{cases} 3y+2x=6\\ 5y-2x=10 \end{cases}

We can eliminate the \(x\)-variable by addition of the two equations.

\begin{cases} 3y+2x=6 \\ \underline{+\: 5y-2x=10} \end{cases}

$$=8y\: \: \: \: \; \; \; \; =16$$

$$\begin{matrix} \: \: \: y\: \: \: \: \: \; \; \; \; \; =2 \end{matrix}$$

The value of \(y\) can now be substituted into either of the original equations to find the value of \(x\)

$$3\cdot {\color{green} 2}+2x=6$$

The solution of the linear system is \((0, 2)\).

To avoid errors make sure that all like terms and equal signs are in the same columns before beginning the elimination.

If you don't have equations where you can eliminate a variable by addition or subtraction you directly you can begin by multiplying one or both of the equations with a constant to obtain an equivalent linear system where you can eliminate one of the variables by addition or subtraction.

\begin{cases} 3x+y=9\\ 5x+4y=22 \end{cases}

Begin by multiplying the first equation by \(-4\) so that the coefficients of \(y\) are opposites

\begin{cases} \color{green}{-4} \cdot (3x + y) = \color{green}{-4} \cdot 9\\ 5x + 4y = 22 \end{cases}

$$\Rightarrow$$

\begin{cases}-12x-4y=-36 \\ \underline{+5x+4y=22 }\end{cases}

$$=-7x\: \: \: \: \: \: \: \: \: \: =-14$$

$$\begin{matrix} \: \:\; \:\: x\: \: \: \: \: \: \: \: \: \: \:=2 \end{matrix}$$

Substitute \(x\) in either of the original equations to get the value of \(y\)

$$3\cdot {\color{green} 2}+y=9$$

The solution of the linear system is \((2, 3)\)

Video lesson

Solve the following linear system using the elimination method

\begin{cases} 2y - 4x = 2 \\ y = -x + 4 \end{cases}

Linear Equations

6.2 Solving Linear Equations

Equations of the form ax+b=0 are called linear equations in the variable x . In this section we will be concerned with the problem of solving linear equations, and equations that reduce to linear equations.

We deﬁne two equations as equivalent if they have the same solution set. The following two operations on an equation always result in a new equation which is equivalent to the original one. These operations, sometimes called elementary transformations, are:

T.1 The same expression representing a real number may be added to both sides of an equation.

T.2 The same expression representing a nonzero real number may be multiplied into both sides of an equation.

Using these operations we may transform an equation whose solution set is not obvious through a series of equivalent equations to an equation that has an obvious solution set.

Example 1. Solve the equation

Add -x to both sides to obtain

-x+2x-3=-x+4+x (T.1)

Add 3 to both sides to obtain

x-3+3=4+3 (T.1)

Since 2x-3=4+x is equivalent to x-3=4 , which, in turn, is equivalent to x=7 , whose solution set is obviously {7} , we know that the solution set of (a) is {7} .

Let’s see how our Linear equation solver solves this and similar problems. Click on "Solve Similar" button to see more examples.

Example 2. Solve the equation

(b) 1/2x+2/3=5/2x-1

Add -(1/2)x to both sides to obtain

2/3=5/2x-1/2x-1 (T.1)

Add 1 to both sides to obtain

Multiply both sides by 1/2 to obtain

Thus, the solution set of (b) is {5/6} .

Every linear equation can be solved in the same way as in the above examples. In fact, let us consider the general linear equation

Add -b to both sides to obtain

Multiply both sides by 1/a to obtain

if a a!=0 . The general linear equation, therefore, has as its solution set {b/a} , if a!=0 . Thus each linear equation has at most one solution.

The next two examples are of equations that reduce to linear equations.

Example 3. Solve the equation

23+4y(5y+4)=9+10y(2y+3)

We expand both sides to obtain

23+20y^2+16y=9+20y^2+30y

Add -20y^2 to both sides to obtain

We now solve as in the previous examples.

Thus the solution set is {1} .

Let’s see how our step by step math solver solves this and similar problems. Click on "Solve Similar" button to see more examples.

Example 4. Solve the equation

(c) (2x)/(x-1)=2/(x-1)+1

The replacement set of (c) is all real numbers except 1. Assuming that x!=1 , we multiply both sides of (c) by x-1 to obtain

(d) 2x=2+x-1,x!=1

Solving the equation 2x =2+x- 1 , we obtain 1 as the only solution Since 1 is not in the replacement of (d), (d) has no solution. Furthermore, (c) is equivalent to (d), therefore (c) has no solution.

6.3 Solving Literal Equations

An equation containing more than one variable, or containing symbols representing constants such as a,b , and c , can be solved for one of the symbols in terms of the remaining symbols by applications of the operations T.1 and T.2 in the preceding section. The student will encounter such problems in other courses.

Example 1. Solve cx-3a=b for x .

Add 3a to both sides.

Multiply both sides by 1/c .

This last equation expresses x in terms of the other symbols.

Example 2. Solve 3ay-2b=2cy for y .

Add 2b to both sides.

Add -2cy to both sides.

Multiply both sides by 1/((3a-2c))

Example 3. Solve a/x+b/(2x)=c for x .

Multiply both sides by 2x .

Multiply by 1/(2c) .

We conclude this section by including two more examples similar to those that the student may encounter in other areas.

Let’s see how our math solver solves this and similar problems. Click on "Solve Similar" button to see more examples.

Example 4. Solve A=P(1+rt) for r .

Apply the distributive law.

Add -P to both sides.

Multiply both sides by 1/(Pt) .

Example 5. Solve 1/R=1/r_1+1/r_2 for r_1 .

Add the two terms on the right—hand side.

1/(R)=(r_2+r_1)/(r_1r_2)

Multiply by Rr_1r_2 .

r_1r_2=R(r_2+r_1)

r_1r_2=Rr_2+Rr_1

Add -Rr_1 to both sides.

r_1r_2-Rr_1=Rr_2

Factor out r_1 .

r_1(r_2-R)=Rr_2

Multiply by 1/(r_2-R) .

r_1=(Rr_2)/(r_2-R)

6.4 Solving Statement Problems

One of the fundamental applications of algebra is solving problems that have been stated in words. A statement problem is a word description of a situation that involves both known and unknown quantities. In this section each problem will be solvable by means of one equation involving one unknown.

Our problem is to choose the unknown and to determine the equation that it must satisfy. Although there is no single approach to all of the problems, the following suggestions are sometimes helpful:

1. Read the problem carefully until the situation is thoroughly understood.

2. Determine what quantities are asked for, then choose the one that seems to be the best to use as the unknown.

3. Establish the relationship between the unknown and the other quantities in the problem.

4. Find the information that tells which two quantities are equal.

5. Use the information in (4) to write the equation.

6. Solve the equation and check the solution to see that it satisﬁes the original problem.

At this stage, the emphasis will be on translating statement problems into equations. Although some of the problems can be solved almost by inspection, the practice that we obtain in setting up equations will prove helpful in working more difficult problems.

Example 1. If 2 times a certain integer is added to the next consecutive integer the result is 34 . Find the integers. Step 1. Reread! Step 2. Let x be the ﬁrst integer. Step 3. Then x+ 1 is the next consecutive integer. Step 4. 2 times a certain integer plus the next consecutive integer is 34 . Step 5. 2x+(x+1)=34

Check. 2*11+(11+1)=34

Example 2. Bob and Joe together earned $ 60 . Both were paid at the same rate, but Bob worked three times as long as Joe. How much did each receive?

Step 1. Reread!

Step 2. Let x be the number of dollars that Joe received.

Step 3. Then 3x is the number of dollars that Bob received

Step 4. Bob and Joe together earned $ 60 . Step 5. 3x+x=60

Check 3*15+15=60

Example 3. The sum of the digits of a two digit number is 12 . If the digits are reversed the number is decreased by 36 . What is the number?

Step 2. Let x be the tens digit.

Step 3. Then 12 - x is the units digit.

Step 4. If the digits are reversed then the number is decreased by 36

Step 5. 10(12-x)+x = 10x+ (12-x) -36

10(12-x)+x = 10x+ (12-x) -36

= 120-10x+x=10x+12-x-36

Therefore the number is 84 .

Check. 84-36=48

Example 4. How many pounds of candy valued at 48 ¢ per pound should be added to 50 pounds of candy valued at 80 ¢ per pound in order for the store owner to be able to sell the candy at 60 ¢ per pound? Step 1. Reread! Step 2. Let x be the number of pounds of 48 ¢ per pound candy. Step 3. Then 50+x would be the pounds of candy he would have at 60 ¢ per pound. Step 4. The amount of candy at 48 ¢ per pound times 48 ¢, plus the amount of candy at 80 ¢ per pound times 80 ¢, must be equal to the amount of candy at 60 ¢ per pound times 60 ¢. Step 5. ( 48 ¢/lb)( x lbs) + ( 80 ¢/lb) ( 50 lbs) = ( 60 ¢/lb) [( 50+x )lbs]

48x+80*50=60(50+x)

48x+4000=3000+60x

x=(83(1)/3) lbs

Check. (83+1/3)48+80*50=60(50+83+1/3)

Problems involving velocities (or speeds) will use the formula

where d is the distance traveled, r is the rate, and t is the time. When the formula is used, d and r must be expressed in the same unit of distance, while r and t must be expressed in the same unit of time.

Example 5. A group of students drove to a lake in the north woods to ﬁsh. They traveled 380 miles in 7 hours, of which 4 hours were on a paved highway and the remaining time was on a dirt road. If the average speed on the dirt road was 25 miles per hour less than the average speed on the highway, then ﬁnd for each part of the trip the average speed and the distance traveled.

Step 1 . Reread!

Step 2. Let x be the speed on the dirt road.

Step 3. Then x+25 is the speed on the highway.

Step 4. The distance traveled on the highway plus the distance traveled on the dirt road is equal to 380 miles.

Step 5. Since d=rt , we have

[(x+25)(mi)/(hr)](4hrs)+[x(mi)/(hr)](3hrs)=380mi

x=40 miles per hour

x+25=65 miles per hour

Check. (40+25)4+40*3=380

Work problems which involve the rate of performance can often be solved by ﬁrst ﬁnding the fractional part of the task done by each person or machine in one unit of time, and then ﬁnding an equation that relates these various fractional parts.

Example 6. A boy can cut a lawn in 4 hours while the father can cut it in 3 hours. How long would it take them to cut the same lawn working together?

Step 2. Let x be the number of hours that it would take them to cut the lawn Working together.

Step 3 . Choose one hour as our unit of time. Now the boy can cut 1/4 of the lawn in one hour, the father can cut 1/3 of the lawn in one hour, and togeﬂner they can cut 1/x of the lawn in one hour.

Step 4. The amount cut by the boy in one hour plus the amount cut by the father in one hour is equal to the amount they can cut together in one hour.

Step 5. 1/3+1/4=1/x

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