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Use a General Strategy to Solve Linear Equations
Solve linear equations using a general strategy
Solve Linear Equations Using a General Strategy
Solving an equation is like discovering the answer to a puzzle. The purpose in solving an equation is to find the value or values of the variable that makes it a true statement. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle!
To determine whether a number is a solution to an equation, we substitute the value for the variable in the equation. If the resulting equation is a true statement, then the number is a solution of the equation.
Example
Try it.
Determine whether the values are solutions to the equation: \(5y+3=10y-4.\)
ⓐ \(y=\frac{3}{5}\) ⓑ \(y=\frac{7}{5}\)
Solution
Since a solution to an equation is a value of the variable that makes the equation true, begin by substituting the value of the solution for the variable.
ⓐ
| Multiply. | |
| Simplify. |
Since \(y=\frac{3}{5}\) does not result in a true equation, \(y=\frac{3}{5}\) is not a solution to the equation \(5y+3=10y-4.\)
ⓑ
| Multiply. | |
| Simplify. |
Since \(y=\frac{7}{5}\) results in a true equation, \(y=\frac{7}{5}\) is a solution to the equation \(5y+3=10y-4.\)
There are many types of equations that we will learn to solve. In this section we will focus on a linear equation.
To solve a linear equation it is a good idea to have an overall strategy that can be used to solve any linear equation. In the next example, we will give the steps of a general strategy for solving any linear equation. Simplifying each side of the equation as much as possible first makes the rest of the steps easier.
How to Solve a Linear Equation Using a General Strategy
Try it.
Solve: \(7(n-3)-8=-15\).
Solution
These steps are summarized in the General Strategy for Solving Linear Equations below.
Condensed — the full section is in OpenStax Intermediate Algebra 2e.
Classify Equations
Whether or not an equation is true depends on the value of the variable. The equation \(7x+8=-13\) is true when we replace the variable, x, with the value \(-3,\) but not true when we replace x with any other value. An equation like this is called a conditional equation. All the equations we have solved so far are conditional equations.
Now let’s consider the equation \(7y+14=7(y+2).\) Do you recognize that the left side and the right side are equivalent? Let’s see what happens when we solve for y.
Solve:
| Distribute. | |
| Subtract \(7y\) to each side to get the \(y’\text{s}\) to one side. | |
| Simplify—the y’s are eliminated. | |
| But \(14=14\) is true. |
This means that the equation \(7y+14=7(y+2)\) is true for any value of y. We say the solution to the equation is all of the real numbers. An equation that is true for any value of the variable is called an identity.
What happens when we solve the equation \(-8z=-8z+9?\)
Solve:
| Add \(8z\) to both sides to leave the constant alone on the right. | |
| Simplify—the \(z’\text{s}\) are eliminated. | |
| But \(0\ne 9.\) |
Solving the equation \(-8z=-8z+9\) led to the false statement \(0=9.\) The equation \(-8z=-8z+9\) will not be true for any value of z. It has no solution. An equation that has no solution, or that is false for all values of the variable, is called a contradiction.
Example
Try it.
Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(6(2n-1)+3=2n-8+5(2n+1).\)
Solution
| Distribute. | |
| Combine like terms. | |
| Subtract \(12n\) from each side to get the n’s to one side. | |
| Simplify. | |
| This is a true statement. | The equation is an identity. |
| The solution is all real numbers. |
Condensed — the full section is in OpenStax Intermediate Algebra 2e.
Solve Equations with Fraction or Decimal Coefficients
We could use the General Strategy to solve the next example. This method would work fine, but many students do not feel very confident when they see all those fractions. So, we are going to show an alternate method to solve equations with fractions. This alternate method eliminates the fractions.
We will apply the Multiplication Property of Equality and multiply both sides of an equation by the least common denominator (LCD) of all the fractions in the equation. The result of this operation will be a new equation, equivalent to the first, but without fractions. This process is called clearing the equation of fractions.
To clear an equation of decimals, we think of all the decimals in their fraction form and then find the LCD of those denominators.
How to Solve Equations with Fraction or Decimal Coefficients
Try it.
Solve: \(\frac{1}{12}x+\frac{5}{6}=\frac{3}{4}.\)
Solution
Notice in the previous example, once we cleared the equation of fractions, the equation was like those we solved earlier in this chapter. We changed the problem to one we already knew how to solve. We then used the General Strategy for Solving Linear Equations.
Example
Try it.
Solve: \(5=\frac{1}{2}y+\frac{2}{3}y-\frac{3}{4}y.\)
Solution
We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.
| Find the LCD of all fractions in the equation. | ||
| The LCD is 12. | ||
| Multiply both sides of the equation by 12. | ||
| Distribute. | ||
| Simplify—notice, no more fractions. | ||
| Combine like terms. | ||
| Divide by five. | ||
| Simplify. | ||
| Check: | ||
| Let \(y=12.\) | ||
In the next example, we’ll distribute before we clear the fractions.
When you multiply both sides of an equation by the LCD of the fractions, make sure you multiply each term by the LCD—even if it does not contain a fraction.
Condensed — the full section is in OpenStax Intermediate Algebra 2e.
Key Concepts
- How to determine whether a number is a solution to an equation
- Substitute the number in for the variable in the equation.
- Simplify the expressions on both sides of the equation.
- Determine whether the resulting equation is true.
If it is true, the number is a solution.
If it is not true, the number is not a solution.
- How to Solve Linear Equations Using a General Strategy
- Simplify each side of the equation as much as possible.
Use the Distributive Property to remove any parentheses.
Combine like terms. - Collect all the variable terms on one side of the equation.
Use the Addition or Subtraction Property of Equality. - Collect all the constant terms on the other side of the equation.
Use the Addition or Subtraction Property of Equality. - Make the coefficient of the variable term equal to 1.
Use the Multiplication or Division Property of Equality.
State the solution to the equation. - Check the solution.
Substitute the solution into the original equation to make sure the result is a true statement.
- Simplify each side of the equation as much as possible.
- How to Solve Equations with Fraction or Decimal Coefficients
- Find the least common denominator (LCD) of all the fractions and decimals (in fraction form) in the equation.
- Multiply both sides of the equation by that LCD. This clears the fractions and decimals.
- Solve using the General Strategy for Solving Linear Equations.
Use a General Strategy to Solve Linear Equations
Solve Equations Using the General Strategy
In the following exercises, determine whether the given values are solutions to the equation.
Try it.
\(6y+10=12y\)
ⓐ \(y=\frac{5}{3}\) ⓑ \(y=-\frac{1}{2}\)
Solution
ⓐ yes ⓑ no
Try it.
\(4x+9=8x\)
ⓐ \(x=-\frac{7}{8}\) ⓑ \(x=\frac{9}{4}\)
Try it.
\(8u-1=6u\)
ⓐ \(u=-\frac{1}{2}\) ⓑ \(u=\frac{1}{2}\)
Solution
ⓐ no ⓑ yes
Try it.
\(9v-2=3v\)
ⓐ \(v=-\frac{1}{3}\) ⓑ \(v=\frac{1}{3}\)
In the following exercises, solve each linear equation.
Try it.
\(15(y-9)=-60\)
Solution
\(y=5\)
Try it.
\(-16(3n+4)=32\)
Try it.
\(\text{-}(w-12)=30\)
Solution
\(w=-18\)
Try it.
\(\text{-}(t-19)=28\)
Try it.
\(51+5(4-q)=56\)
Solution
\(q=3\)
Try it.
\(-6+6(5-k)=15\)
Try it.
\(3(10-2x)+54=0\)
Solution
\(x=14\)
Try it.
\(-2(11-7x)+54=4\)
Try it.
\(\frac{2}{3}(9c-3)=22\)
Solution
\(c=4\)
Try it.
\(\frac{3}{5}(10x-5)=27\)
Try it.
\(\frac{1}{5}(15c+10)=c+7\)
Solution
\(c=\frac{5}{2}\)
Try it.
\(\frac{1}{4}(20d+12)=d+7\)
Try it.
\(3(4n-1)-2=8n+3\)
Solution
\(n=2\)
Try it.
\(9(2m-3)-8=4m+7\)
Try it.
\(12+2(5-3y)=-9(y-1)-2\)
Solution
\(y=-5\)
Try it.
\(-15+4(2-5y)=-7(y-4)+4\)
Try it.
\(5+6(3s-5)=-3+2(8s-1)\)
Solution
\(s=10\)
Try it.
\(-12+8(x-5)=-4+3(5x-2)\)
Try it.
\(4(p-4)-(p+7)=5(p-3)\)
Solution
\(p=-4\)
Try it.
\(3(a-2)-(a+6)=4(a-1)\)
Try it.
\(4[5-8(4c-3)]=12(1-13c)-8\)
Solution
\(c=-4\)
Try it.
\(5[9-2(6d-1)]=11(4-10d)-139\)
Try it.
\(3[-9+8(4h-3)]=2(5-12h)-19\)
Solution
\(h=\frac{3}{4}\)
Try it.
\(3[-14+2(15k-6)]=8(3-5k)-24\)
Try it.
\(5[2(m+4)+8(m-7)]=2[3(5+m)-(21-3m)]\)
Solution
\(m=6\)
Try it.
\(10[5(n+1)+4(n-1)]=11[7(5+n)-(25-3n)]\)
Classify Equations
In the following exercises, classify each equation as a conditional equation, an identity, or a contradiction and then state the solution.
Try it.
\(23z+19=3(5z-9)+8z+46\)
Solution
identity; all real numbers
Try it.
\(15y+32=2(10y-7)-5y+46\)
Try it.
\(18(5j-1)+29=47\)
Solution
conditional equation;\(j=\frac{2}{5}\)
Try it.
\(24(3d-4)+100=52\)
Try it.
\(22(3m-4)=8(2m+9)\)
Solution
conditional equation; \(m=\frac{16}{5}\)
Try it.
\(30(2n-1)=5(10n+8)\)
Try it.
\(7v+42=11(3v+8)-2(13v-1)\)
Solution
contradiction; no solution
Try it.
\(18u-51=9(4u+5)-6(3u-10)\)
Try it.
\(45(3y-2)=9(15y-6)\)
Solution
contradiction; no solution
Try it.
\(60(2x-1)=15(8x+5)\)
Try it.
\(9(14d+9)+4d=13(10d+6)+3\)
Solution
identity; all real numbers
Try it.
\(11(8c+5)-8c=2(40c+25)+5\)
Solve Equations with Fraction or Decimal Coefficients
In the following exercises, solve each equation with fraction coefficients.
Try it.
\(\frac{1}{4}x-\frac{1}{2}=-\frac{3}{4}\)
Solution
\(x=-1\)
Try it.
\(\frac{3}{4}x-\frac{1}{2}=\frac{1}{4}\)
Try it.
\(\frac{5}{6}y-\frac{2}{3}=-\frac{3}{2}\)
Solution
\(y=-1\)
Try it.
\(\frac{5}{6}y-\frac{1}{3}=-\frac{7}{6}\)
Try it.
\(\frac{1}{2}a+\frac{3}{8}=\frac{3}{4}\)
Solution
\(a=\frac{3}{4}\)
Try it.
\(\frac{5}{8}b+\frac{1}{2}=-\frac{3}{4}\)
Try it.
\(2=\frac{1}{3}x-\frac{1}{2}x+\frac{2}{3}x\)
Solution
\(x=4\)
Try it.
\(2=\frac{3}{5}x-\frac{1}{3}x+\frac{2}{5}x\)
Try it.
\(\frac{1}{3}w+\frac{5}{4}=w-\frac{1}{4}\)
Solution
\(w=\frac{9}{4}\)
Try it.
\(\frac{1}{2}a-\frac{1}{4}=\frac{1}{6}a+\frac{1}{12}\)
Try it.
\(\frac{1}{3}b+\frac{1}{5}=\frac{2}{5}b-\frac{3}{5}\)
Solution
\(b=12\)
Try it.
\(\frac{1}{3}x+\frac{2}{5}=\frac{1}{5}x-\frac{2}{5}\)
Try it.
\(\frac{1}{4}(p-7)=\frac{1}{3}(p+5)\)
Solution
\(p=-41\)
Try it.
\(\frac{1}{5}(q+3)=\frac{1}{2}(q-3)\)
Try it.
\(\frac{1}{2}(x+4)=\frac{3}{4}\)
Solution
\(x=-\frac{5}{2}\)
Try it.
\(\frac{1}{3}(x+5)=\frac{5}{6}\)
Try it.
\(\frac{4n+8}{4}=\frac{n}{3}\)
Solution
\(n=-3\)
Try it.
\(\frac{3p+6}{3}=\frac{p}{2}\)
Try it.
\(\frac{3x+4}{2}+1=\frac{5x+10}{8}\)
Solution
\(x=-2\)
Try it.
\(\frac{10y-2}{3}+3=\frac{10y+1}{9}\)
Try it.
\(\frac{7u-1}{4}-1=\frac{4u+8}{5}\)
Solution
\(u=3\)
Try it.
\(\frac{3v-6}{2}+5=\frac{11v-4}{5}\)
In the following exercises, solve each equation with decimal coefficients.
Try it.
\(0.4x+0.6=0.5x-1.2\)
Solution
\(x=18\)
Try it.
\(0.7x+0.4=0.6x+2.4\)
Try it.
\(0.9x-1.25=0.75x+1.75\)
Solution
\(x=20\)
Try it.
\(1.2x-0.91=0.8x+2.29\)
Try it.
\(0.05n+0.10(n+8)=2.15\)
Solution
\(n=9\)
Try it.
\(0.05n+0.10(n+7)=3.55\)
Try it.
\(0.10d+0.25(d+5)=4.05\)
Solution
\(d=8\)
Try it.
\(0.10d+0.25(d+7)=5.25\)
Condensed — the full section is in OpenStax Intermediate Algebra 2e.
Solve Equations Using the General Strategy
Until now we have dealt with solving one specific form of a linear equation. It is time now to lay out one overall strategy that can be used to solve any linear equation. Some equations we solve will not require all these steps to solve, but many will.
Beginning by simplifying each side of the equation makes the remaining steps easier.
How to Solve Linear Equations Using the General Strategy
Try it.
Solve: \(-6(x+3)=24.\)
Solution
Example
Try it.
Solve: \(\text{-}(y+9)=8.\)
Solution
| Simplify each side of the equation as much as possible by distributing. | ||
| The only \(y\) term is on the left side, so all variable terms are on the left side of the equation. | ||
| Add \(9\) to both sides to get all constant terms on the right side of the equation. | ||
| Simplify. | ||
| Rewrite \(-y\) as \(-1y\). | ||
| Make the coefficient of the variable term to equal to \(1\) by dividing both sides by \(-1\). | ||
| Simplify. | ||
| Check: | ||
| Let \(y=-17\). | ||
Example
Try it.
Solve: \(5(a-3)+5=-10\).
Solution
| Simplify each side of the equation as much as possible. | ||
| Distribute. | ||
| Combine like terms. | ||
| The only \(a\) term is on the left side, so all variable terms are on one side of the equation. | ||
| Add \(10\) to both sides to get all constant terms on the other side of the equation. | ||
| Simplify. | ||
| Make the coefficient of the variable term to equal to \(1\) by dividing both sides by \(5\). | ||
| Simplify. | ||
| Check: | ||
| Let \(a=0\). | ||
Example
Try it.
Solve: \(\frac{2}{3}(6m-3)=8-m\).
Solution
| Distribute. | ||
| Add \(m\) to get the variables only to the left. | ||
| Simplify. | ||
| Add \(2\) to get constants only on the right. | ||
| Simplify. | ||
| Divide by \(5\). | ||
| Simplify. | ||
| Check: | ||
| Let \(m=2\). | ||
Condensed — the full section is in OpenStax Elementary Algebra 2e.
Classify Equations
Consider the equation we solved at the start of the last section, \(7x+8=-13\). The solution we found was \(x=-3\). This means the equation \(7x+8=-13\) is true when we replace the variable, x, with the value \(-3\). We showed this when we checked the solution \(x=-3\) and evaluated \(7x+8=-13\) for \(x=-3\).
If we evaluate \(7x+8\) for a different value of x, the left side will not be \(-13\).
The equation \(7x+8=-13\) is true when we replace the variable, x, with the value \(-3\), but not true when we replace x with any other value. Whether or not the equation \(7x+8=-13\) is true depends on the value of the variable. Equations like this are called conditional equations.
All the equations we have solved so far are conditional equations.
Now let’s consider the equation \(2y+6=2(y+3)\). Do you recognize that the left side and the right side are equivalent? Let’s see what happens when we solve for y.
| Distribute. | |
| Subtract \(2y\) to get the \(y\)’s to one side. | |
| Simplify—the \(y\)’s are gone! |
But \(6=6\) is true.
This means that the equation \(2y+6=2(y+3)\) is true for any value of y. We say the solution to the equation is all of the real numbers. An equation that is true for any value of the variable like this is called an identity.
| Subtract \(5z\) to get the constant alone on the right. | |
| Simplify—the \(z\)’s are gone! |
Example
Try it.
Classify the equation as a conditional equation, an identity, or a contradiction. Then state the solution.
\(6(2n-1)+3=2n-8+5(2n+1)\)
Solution
| Distribute. | |
| Combine like terms. | |
| Subtract \(12n\) to get the \(n\)’s to one side. | |
| Simplify. | |
| This is a true statement. | The equation is an identity. The solution is all real numbers. |
Condensed — the full section is in OpenStax Elementary Algebra 2e.
Key Concepts
- General Strategy for Solving Linear Equations
- Simplify each side of the equation as much as possible.
Use the Distributive Property to remove any parentheses.
Combine like terms. - Collect all the variable terms on one side of the equation.
Use the Addition or Subtraction Property of Equality. - Collect all the constant terms on the other side of the equation.
Use the Addition or Subtraction Property of Equality. - Make the coefficient of the variable term to equal to 1.
Use the Multiplication or Division Property of Equality.
State the solution to the equation. - Check the solution.
Substitute the solution into the original equation.
- Simplify each side of the equation as much as possible.
Use a General Strategy to Solve Linear Equations
Solve Equations Using the General Strategy for Solving Linear Equations
In the following exercises, solve each linear equation.
Try it.
\(15(y-9)=-60\)
Try it.
\(21(y-5)=-42\)
Solution
\(y=3\)
Try it.
\(-9(2n+1)=36\)
Try it.
\(-16(3n+4)=32\)
Solution
\(n=-2\)
Try it.
\(8(22+11r)=0\)
Try it.
\(5(8+6p)=0\)
Solution
\(p=-\frac{4}{3}\)
Try it.
\(\text{-}(w-12)=30\)
Try it.
\(\text{-}(t-19)=28\)
Solution
\(t=-9\)
Try it.
\(9(6a+8)+9=81\)
Try it.
\(8(9b-4)-12=100\)
Solution
\(b=2\)
Try it.
\(32+3(z+4)=41\)
Try it.
\(21+2(m-4)=25\)
Solution
\(m=6\)
Try it.
\(51+5(4-q)=56\)
Try it.
\(-6+6(5-k)=15\)
Solution
\(k=\frac{3}{2}\)
Try it.
\(2(9s-6)-62=16\)
Try it.
\(8(6t-5)-35=-27\)
Solution
\(t=1\)
Try it.
\(3(10-2x)+54=0\)
Try it.
\(-2(11-7x)+54=4\)
Solution
\(x=-2\)
Try it.
\(\frac{2}{3}(9c-3)=22\)
Try it.
\(\frac{3}{5}(10x-5)=27\)
Solution
\(x=5\)
Try it.
\(\frac{1}{5}(15c+10)=c+7\)
Try it.
\(\frac{1}{4}(20d+12)=d+7\)
Solution
\(d=1\)
Try it.
\(18-(9r+7)=-16\)
Try it.
\(15-(3r+8)=28\)
Solution
\(r=-7\)
Try it.
\(5-(n-1)=19\)
Try it.
\(-3-(m-1)=13\)
Solution
\(m=-15\)
Try it.
\(11-4(y-8)=43\)
Try it.
\(18-2(y-3)=32\)
Solution
\(y=-4\)
Try it.
\(24-8(3v+6)=0\)
Try it.
\(35-5(2w+8)=-10\)
Solution
\(w=\frac{1}{2}\)
Try it.
\(4(a-12)=3(a+5)\)
Try it.
\(-2(a-6)=4(a-3)\)
Solution
\(a=4\)
Try it.
\(2(5-u)=-3(2u+6)\)
Try it.
\(5(8-r)=-2(2r-16)\)
Solution
\(r=8\)
Try it.
\(3(4n-1)-2=8n+3\)
Try it.
\(9(2m-3)-8=4m+7\)
Solution
\(m=3\)
Try it.
\(12+2(5-3y)=-9(y-1)-2\)
Try it.
\(-15+4(2-5y)=-7(y-4)+4\)
Solution
\(y=-3\)
Try it.
\(8(x-4)-7x=14\)
Try it.
\(5(x-4)-4x=14\)
Solution
\(x=34\)
Try it.
\(5+6(3s-5)=-3+2(8s-1)\)
Try it.
\(-12+8(x-5)=-4+3(5x-2)\)
Solution
\(x=-6\)
Try it.
\(4(u-1)-8=6(3u-2)-7\)
Try it.
\(7(2n-5)=8(4n-1)-9\)
Solution
\(n=-1\)
Try it.
\(4(p-4)-(p+7)=5(p-3)\)
Try it.
\(3(a-2)-(a+6)=4(a-1)\)
Solution
\(a=-4\)
Try it.
\(\text{-}(9y+5)-(3y-7)\)
\(=16-(4y-2)\)
Try it.
\(\text{-}(7m+4)-(2m-5)\)
\(=14-(5m-3)\)
Solution
\(m=-4\)
Try it.
\(4[5-8(4c-3)]\)
\(=12(1-13c)-8\)
Try it.
\(5[9-2(6d-1)]\)
\(=11(4-10d)-139\)
Solution
\(d=-3\)
Try it.
\(3[-9+8(4h-3)]\)
\(=2(5-12h)-19\)
Try it.
\(3[-14+2(15k-6)]\)
\(=8(3-5k)-24\)
Solution
\(k=\frac{3}{5}\)
Try it.
\(5[2(m+4)+8(m-7)]\)
\(=2[3(5+m)-(21-3m)]\)
Try it.
\(10[5(n+1)+4(n-1)]\)
\(=11[7(5+n)-(25-3n)]\)
Solution
\(n=-5\)
Try it.
\(5(1.2u-4.8)=-12\)
Try it.
\(4(2.5v-0.6)=7.6\)
Solution
\(v=1\)
Try it.
\(0.25(q-6)=0.1(q+18)\)
Try it.
\(0.2(p-6)=0.4(p+14)\)
Solution
\(p=-34\)
Try it.
\(0.2(30n+50)=28\)
Try it.
\(0.5(16m+34)=-15\)
Solution
\(m=-4\)
Classify Equations
In the following exercises, classify each equation as a conditional equation, an identity, or a contradiction and then state the solution.
Try it.
\(23z+19=3(5z-9)+8z+46\)
Try it.
\(15y+32=2(10y-7)-5y+46\)
Solution
identity; all real numbers
Try it.
\(5(b-9)+4(3b+9)=6(4b-5)-7b+21\)
Try it.
\(9(a-4)+3(2a+5)=7(3a-4)-6a+7\)
Solution
identity; all real numbers
Try it.
\(18(5j-1)+29=47\)
Try it.
\(24(3d-4)+100=52\)
Solution
conditional equation; \(d=\frac{2}{3}\)
Try it.
\(22(3m-4)=8(2m+9)\)
Try it.
\(30(2n-1)=5(10n+8)\)
Solution
conditional equation; \(n=7\)
Try it.
\(7v+42=11(3v+8)-2(13v-1)\)
Try it.
\(18u-51=9(4u+5)-6(3u-10)\)
Solution
contradiction; no solution
Try it.
\(3(6q-9)+7(q+4)=5(6q+8)-5(q+1)\)
Try it.
\(5(p+4)+8(2p-1)=9(3p-5)-6(p-2)\)
Solution
contradiction; no solution
Try it.
\(12(6h-1)=8(8h+5)-4\)
Try it.
\(9(4k-7)=11(3k+1)+4\)
Solution
conditional equation; \(k=26\)
Try it.
\(45(3y-2)=9(15y-6)\)
Try it.
\(60(2x-1)=15(8x+5)\)
Solution
contradiction; no solution
Try it.
\(16(6n+15)=48(2n+5)\)
Try it.
\(36(4m+5)=12(12m+15)\)
Solution
identity; all real numbers
Try it.
\(9(14d+9)+4d=13(10d+6)+3\)
Try it.
\(11(8c+5)-8c=2(40c+25)+5\)
Solution
identity; all real numbers
Condensed — the full section is in OpenStax Elementary Algebra 2e.
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
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Simplify: \(\frac{3}{2}(12x+20).\)
If you missed this problem, review .जवाफ प्रकट गर्नुहोस्
\(18x+30\)
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Simplify: \(5-2(n+1).\)
If you missed this problem, review .जवाफ प्रकट गर्नुहोस्
\(3-2n\)
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Find the LCD of \(\frac{5}{6}\) and \(\frac{1}{4}.\)
If you missed this problem, review .जवाफ प्रकट गर्नुहोस्
\(12\)
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Determine whether the values are solutions to the equation: \(5y+3=10y-4.\)
ⓐ \(y=\frac{3}{5}\) ⓑ \(y=\frac{7}{5}\)
जवाफ प्रकट गर्नुहोस्
Since a solution to an equation is a value of the variable that makes the equation true, begin by substituting the value of the solution for the variable.
ⓐ
Multiply. Simplify. Since \(y=\frac{3}{5}\) does not result in a true equation, \(y=\frac{3}{5}\) is not a solution to the equation \(5y+3=10y-4.\)
ⓑ
Multiply. Simplify. Since \(y=\frac{7}{5}\) results in a true equation, \(y=\frac{7}{5}\) is a solution to the equation \(5y+3=10y-4.\)
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Determine whether the values are solutions to the equation: \(9y+2=6y+3.\)
ⓐ \(y=\frac{4}{3}\) ⓑ \(y=\frac{1}{3}\)
जवाफ प्रकट गर्नुहोस्
ⓐ no ⓑ yes
-
Determine whether the values are solutions to the equation: \(4x-2=2x+1.\)
ⓐ \(x=\frac{3}{2}\) ⓑ \(x=-\frac{1}{2}\)
जवाफ प्रकट गर्नुहोस्
ⓐ yes ⓑ no
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Solve: \(7(n-3)-8=-15\).
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Solve: \(2(m-4)+3=-1.\)
जवाफ प्रकट गर्नुहोस्
\(m=2\)
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Solve: \(5(a-3)+5=-10.\)
जवाफ प्रकट गर्नुहोस्
\(a=0\)
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Solve: \(\frac{2}{3}(3m-6)=5-m.\)
जवाफ प्रकट गर्नुहोस्
Distribute. Add m to both sides to get the variables only on the left. Simplify. Add 4 to both sides to get constants only on the right. Simplify. Divide both sides by three. Simplify. Check: Let \(m=3.\) -
Solve: \(\frac{1}{3}(6u+3)=7-u.\)
जवाफ प्रकट गर्नुहोस्
\(u=2\)
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Solve: \(\frac{2}{3}(9x-12)=8+2x.\)
जवाफ प्रकट गर्नुहोस्
\(x=4\)
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Solve: \(4(x-1)-2=5(2x+3)+6.\)
जवाफ प्रकट गर्नुहोस्
Distribute. Combine like terms. Subtract \(4x\) from each side to get the variables only on
the right since \(10>4.\)Simplify. Subtract 21 from each side to get the constants on left. Simplify. Divide both sides by 6. Simplify. Check: Let \(x=-\frac{9}{2}.\) -
Solve: \(6(p-3)-7=5(4p+3)-12.\)
जवाफ प्रकट गर्नुहोस्
\(p=-2\)
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Solve: \(8(q+1)-5=3(2q-4)-1.\)
जवाफ प्रकट गर्नुहोस्
\(q=-8\)
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Solve: \(10[3-8(2s-5)]=15(40-5s).\)
जवाफ प्रकट गर्नुहोस्
Simplify from the innermost parentheses first. Combine like terms in the brackets. Distribute. Add \(160s\) to both sides to get the
variables to the right.Simplify. Subtract 600 from both sides to get the
constants to the left.Simplify. Divide both sides by 85. Simplify. Check: Let \(s=-2.\) -
Solve: \(6[4-2(7y-1)]=8(13-8y).\)
जवाफ प्रकट गर्नुहोस्
\(y=-\frac{17}{5}\)
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Solve: \(12[1-5(4z-1)]=3(24+11z).\)
जवाफ प्रकट गर्नुहोस्
\(z=0\)
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Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(6(2n-1)+3=2n-8+5(2n+1).\)
जवाफ प्रकट गर्नुहोस्
Distribute. Combine like terms. Subtract \(12n\) from each side to get the n’s to one side. Simplify. This is a true statement. The equation is an identity. The solution is all real numbers. -
Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(4+9(3x-7)=-42x-13+23(3x-2).\)
जवाफ प्रकट गर्नुहोस्
identity; all real numbers
-
Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(8(1-3x)+15(2x+7)=2(x+50)+4(x+3)+1.\)
जवाफ प्रकट गर्नुहोस्
identity; all real numbers
-
Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(8+3(a-4)=0.\)
जवाफ प्रकट गर्नुहोस्
Distribute. Combine like terms. Add 4 to both sides. Simplify. Divide. Simplify. The equation is true when \(a=\frac{4}{3}.\\) This is a conditional equation. The solution is \(a=\frac{4}{3}.\) -
Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(11(q+3)-5=19.\)
जवाफ प्रकट गर्नुहोस्
conditional equation; \(q=-\frac{9}{11}\)
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Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(6+14(k-8)=95.\)
जवाफ प्रकट गर्नुहोस्
conditional equation; \(k=\frac{201}{14}\)
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Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(5m+3(9+3m)=2(7m-11).\)
जवाफ प्रकट गर्नुहोस्
Distribute. Combine like terms. Subtract \(14m\) from both sides. Simplify. But \(27\ne \text{-}22.\) The equation is a contradiction. It has no solution. -
Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(12c+5(5+3c)=3(9c-4).\)
जवाफ प्रकट गर्नुहोस्
contradiction; no solution
-
Classify the equation as a conditional equation, an identity, or a contradiction and then state the solution: \(4(7d+18)=13(3d-2)-11d.\)
जवाफ प्रकट गर्नुहोस्
contradiction; no solution
-
Solve: \(\frac{1}{12}x+\frac{5}{6}=\frac{3}{4}.\)
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Solve: \(\frac{1}{4}x+\frac{1}{2}=\frac{5}{8}.\)
जवाफ प्रकट गर्नुहोस्
\(x=\frac{1}{2}\)
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Solve: \(\frac{1}{8}x+\frac{1}{2}=\frac{1}{4}.\)
जवाफ प्रकट गर्नुहोस्
\(x=-2\)
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Solve: \(5=\frac{1}{2}y+\frac{2}{3}y-\frac{3}{4}y.\)
जवाफ प्रकट गर्नुहोस्
We want to clear the fractions by multiplying both sides of the equation by the LCD of all the fractions in the equation.
Find the LCD of all fractions in the equation. The LCD is 12. Multiply both sides of the equation by 12. Distribute. Simplify—notice, no more fractions. Combine like terms. Divide by five. Simplify. Check: Let \(y=12.\) -
Solve: \(7=\frac{1}{2}x+\frac{3}{4}x-\frac{2}{3}x.\)
जवाफ प्रकट गर्नुहोस्
\(x=12\)
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Solve: \(-1=\frac{1}{2}u+\frac{1}{4}u-\frac{2}{3}u.\)
जवाफ प्रकट गर्नुहोस्
\(u=-12\)
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Solve: \(\frac{1}{2}(y-5)=\frac{1}{4}(y-1).\)
जवाफ प्रकट गर्नुहोस्
Distribute. Simplify. Multiply by the LCD, four. Distribute. Simplify. Collect the variables to the left. Simplify. Collect the constants to the right. Simplify. An alternate way to solve this equation is to clear the fractions without distributing first. If you multiply the factors correctly, this method will be easier. Multiply by the LCD, 4. Multiply four times the fractions. Distribute. Collect the variables to the left. Simplify. Collect the constants to the right. Simplify. Check: Let \(y=9.\) Finish the check on your own. -
Solve: \(\frac{1}{5}(n+3)=\frac{1}{4}(n+2).\)
जवाफ प्रकट गर्नुहोस्
\(n=2\)
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Solve: \(\frac{1}{2}(m-3)=\frac{1}{4}(m-7).\)
जवाफ प्रकट गर्नुहोस्
\(m=-1\)
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Solve: \(\frac{4q+3}{2}+6=\frac{3q+5}{4}\)
जवाफ प्रकट गर्नुहोस्
Multiply both sides by the LCD, 4. Distribute. Simplify. Collect the variables to the left. Simplify. Collect the constants to the right. Simplify. Divide both sides by five. Simplify. Check: Let \(q=-5.\) Finish the check on your own. -
Solve: \(\frac{3r+5}{6}+1=\frac{4r+3}{3}.\)
जवाफ प्रकट गर्नुहोस्
\(r=1\)
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Solve: \(\frac{2s+3}{2}+1=\frac{3s+2}{4}.\)
जवाफ प्रकट गर्नुहोस्
\(s=-8\)
-
Solve: \(0.25x+0.05(x+3)=2.85.\)
जवाफ प्रकट गर्नुहोस्
Look at the decimals and think of the equivalent fractions:
\[0.25=\frac{25}{100},\ 0.05=\frac{5}{100},\ 2.85=2\frac{85}{100}.\]Notice, the LCD is 100. By multiplying by the LCD we will clear the decimals from the equation.
Distribute first. Combine like terms. To clear decimals, multiply by 100. Distribute. Subtract 15 from both sides. Simplify. Divide by 30. Simplify. Check it yourself by substituting \(x=9\) into the original equation.
Symbols used here
The two sides are different.
Both signs at once: x = 3 ± 2 means 5 and 1.
Inequalities that allow equality; < and > exclude it.
The non-negative number whose square (n-th power) is x.
Distance from zero: |−3| = 3. For a complex number, distance from the origin.
i² = −1.
The exponent b must be raised to for x; ln uses base e.
Naturals, integers, rationals, reals, complex numbers.
How to: Use a General Strategy to Solve Linear Equations
- Solve linear equations using a general strategy
- Classify equations
- Solve equations with fraction or decimal coefficients
- Substitute the number for the variable in the equation.
- Simplify the expressions on both sides of the equation.
- Determine whether the resulting equation is true.
- If it is true, the number is a solution.
- If it is not true, the number is not a solution.
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.
तपाईँको आफ्नै प्रयास गर्नुहोस्
Parts of this page are adapted from OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0), OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
यसमा थप Algebra
Linear equationsQuadratic equationsSystems of equationsInequalitiesFactoringExpandingSimplifying expressionsFunctions and graphsExponential and logarithmic equationsPolynomial equationsAbsolute value