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Solve: 6 n = 5 n 10 .

n = −10

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Solve: −6 c = −7 c 1 .

c = −1

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Solve: 5 y 9 = 8 y .

Solution

The only constant is on the left and the y ’s are on both sides. Let’s leave the constant on the left and get the variables to the right.

.
Subtract 5 y from both sides. .
Simplify. .
We have the y ’s on the right and the
constants on the left. Divide both sides by 3.
.
Simplify. .
Check: .
Let y = −3 . .
.
.

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Solve: 3 p 14 = 5 p .

p = −7

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Solve: 8 m + 9 = 5 m .

m = −3

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Solve: 12 x = x + 26 .

Solution

The only constant is on the right, so let the left side be the “variable” side.

.
Remove the x from the right side by adding x to both sides. .
Simplify. .
All the x ’s are on the left and the constants are on the right. Divide both sides by 13. .
Simplify. .
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Solve: 12 j = −4 j + 32 .

j = 2

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Solve: 8 h = −4 h + 12 .

h = 1

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Solve equations with variables and constants on both sides

The next example will be the first to have variables and constants on both sides of the equation. It may take several steps to solve this equation, so we need a clear and organized strategy.

How to solve equations with variables and constants on both sides

Solve: 7 x + 5 = 6 x + 2 .

Solution

This figure is a table that has three columns and four rows. The first column is a header column, and it contains the names and numbers of each step. The second column contains further written instructions. The third column contains math. On the top row of the table, the first cell on the left reads: “Step 1. Choose which side will the “variable” side—the other side will be the “constant” side.” The text in the second cell reads: “The variable terms are 7 x and 6 x. Since 7 is greater than 6, we will make the left side the “x” side and so the right side will be the “constant” side.” The third cell contains the equation 7 x plus 5 equals 6 x plus 2, and the left side of the equation is labeled “variable” written in red, and the right side of the equation is labeled “constant” written in red. In the second row of the table, the first cell says: “Step 2. Collect the variable terms to the “variable” side of the equation, using the addition or subtraction property of equality.” In the second cell, the instructions say: “ With the right side as the “constant” side, the 6x is out of place, so subtract 6x from both sides. Combine like terms. Now the variable is only on the left side!” The third cell contains the original equation with 6x subtracted from both sides: 7 x minus 6 x plus 5 equals 6 x minus 6 x plus 2, with “minus 6 x” written in red on both sides. Below this is the same equation with like terms combined: x plus 5 equals 2. In the third row of the table, the first cell says: “Step 3. Collect all the constants to the other side of the equation, using the addition or subtraction property of equality.” In the second cell, the instructions say: “The right side is the “constant” side, so the 5 is out of place. Subtract 5 from both sides. Simplify.” The third cell contains the equation x plus 5 minus 5 equals 2 minus 5, with “minus 5” written in red on both sides. Below this is the answer to the equation: x equals negative 3. In the fourth row of the table, the first cell says: “Step 4. Make the coefficient of the variable equal 1, using the Multiplication or Division Property of Equality.” In the second cell, the instructions say: “The coefficient of x is one. The equation is solved.” The third cell is blank. In the fifth row of the table, the first cell says: “Step 5. Check.” The instructions in the second cell say: “Check. Let x equal negative 3. Simplify. Add.” In the third cell is the original equation again: 7 x plus 5 equals 6x plus 2. Below this is the same equation with negative 3 substituted in for x: 7 times negative 3 (in paretheses) plus 5 might equal 6 times negative 3 (in parentheses) plus 2, with the “times negative 3” written in red on both sides of the equation. Below this is the equation negative 21 plus 5 might equal negative 18 plus 2. On the last line is the equation negative 16 equals negative 16, with a check mark next to it.
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Solve: 12 x + 8 = 6 x + 2 .

x = −1

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Solve: 9 y + 4 = 7 y + 12 .

y = 4

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We’ll list the steps below so you can easily refer to them. But we’ll call this the ‘Beginning Strategy’ because we’ll be adding some steps later in this chapter.

Beginning strategy for solving equations with variables and constants on both sides of the equation.

  1. Choose which side will be the “variable” side—the other side will be the “constant” side.
  2. Collect the variable terms to the “variable” side of the equation, using the Addition or Subtraction Property of Equality.
  3. Collect all the constants to the other side of the equation, using the Addition or Subtraction Property of Equality.
  4. Make the coefficient of the variable equal 1, using the Multiplication or Division Property of Equality.
  5. Check the solution by substituting it into the original equation.

In Step 1, a helpful approach is to make the “variable” side the side that has the variable with the larger coefficient. This usually makes the arithmetic easier.

Solve: 8 n 4 = −2 n + 6 .

Solution

In the first step, choose the variable side by comparing the coefficients of the variables on each side.

Since 8 > −2 , make the left side the “variable” side. .
We don’t want variable terms on the right side—add 2 n to both sides to leave only constants on the right. .
Combine like terms. .
We don’t want any constants on the left side, so add 4 to both sides. .
Simplify. .
The variable term is on the left and the constant term is on the right. To get the coefficient of n to be one, divide both sides by 10. .
Simplify. .
Check: .
Let n = 1 . .
.
.
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Solve: 8 q 5 = −4 q + 7 .

q = 1

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Solve: 7 n 3 = n + 3 .

n = 1

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Solve: 7 a 3 = 13 a + 7 .

Solution

In the first step, choose the variable side by comparing the coefficients of the variables on each side.

Since 13 > 7 , make the right side the “variable” side and the left side the “constant” side.

.
Subtract 7 a from both sides to remove the variable term from the left. .
Combine like terms. .
Subtract 7 from both sides to remove the constant from the right. .
Simplify. .
Divide both sides by 6 to make 1 the coefficient of a . .
Simplify. .
Check: .
Let a = 5 3 . .
.
.
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Questions & Answers

Three charges q_{1}=+3\mu C, q_{2}=+6\mu C and q_{3}=+8\mu C are located at (2,0)m (0,0)m and (0,3) coordinates respectively. Find the magnitude and direction acted upon q_{2} by the two other charges.Draw the correct graphical illustration of the problem above showing the direction of all forces.
Kate Reply
To solve this problem, we need to first find the net force acting on charge q_{2}. The magnitude of the force exerted by q_{1} on q_{2} is given by F=\frac{kq_{1}q_{2}}{r^{2}} where k is the Coulomb constant, q_{1} and q_{2} are the charges of the particles, and r is the distance between them.
Muhammed
What is the direction and net electric force on q_{1}= 5µC located at (0,4)r due to charges q_{2}=7mu located at (0,0)m and q_{3}=3\mu C located at (4,0)m?
Kate Reply
what is the change in momentum of a body?
Eunice Reply
what is a capacitor?
Raymond Reply
Capacitor is a separation of opposite charges using an insulator of very small dimension between them. Capacitor is used for allowing an AC (alternating current) to pass while a DC (direct current) is blocked.
Gautam
A motor travelling at 72km/m on sighting a stop sign applying the breaks such that under constant deaccelerate in the meters of 50 metres what is the magnitude of the accelerate
Maria Reply
please solve
Sharon
8m/s²
Aishat
What is Thermodynamics
Muordit
velocity can be 72 km/h in question. 72 km/h=20 m/s, v^2=2.a.x , 20^2=2.a.50, a=4 m/s^2.
Mehmet
A boat travels due east at a speed of 40meter per seconds across a river flowing due south at 30meter per seconds. what is the resultant speed of the boat
Saheed Reply
50 m/s due south east
Someone
which has a higher temperature, 1cup of boiling water or 1teapot of boiling water which can transfer more heat 1cup of boiling water or 1 teapot of boiling water explain your . answer
Ramon Reply
I believe temperature being an intensive property does not change for any amount of boiling water whereas heat being an extensive property changes with amount/size of the system.
Someone
Scratch that
Someone
temperature for any amount of water to boil at ntp is 100⁰C (it is a state function and and intensive property) and it depends both will give same amount of heat because the surface available for heat transfer is greater in case of the kettle as well as the heat stored in it but if you talk.....
Someone
about the amount of heat stored in the system then in that case since the mass of water in the kettle is greater so more energy is required to raise the temperature b/c more molecules of water are present in the kettle
Someone
definitely of physics
Haryormhidey Reply
how many start and codon
Esrael Reply
what is field
Felix Reply
physics, biology and chemistry this is my Field
ALIYU
field is a region of space under the influence of some physical properties
Collete
what is ogarnic chemistry
WISDOM Reply
determine the slope giving that 3y+ 2x-14=0
WISDOM
Another formula for Acceleration
Belty Reply
a=v/t. a=f/m a
IHUMA
innocent
Adah
pratica A on solution of hydro chloric acid,B is a solution containing 0.5000 mole ofsodium chlorid per dm³,put A in the burret and titrate 20.00 or 25.00cm³ portion of B using melting orange as the indicator. record the deside of your burret tabulate the burret reading and calculate the average volume of acid used?
Nassze Reply
how do lnternal energy measures
Esrael
Two bodies attract each other electrically. Do they both have to be charged? Answer the same question if the bodies repel one another.
JALLAH Reply
No. According to Isac Newtons law. this two bodies maybe you and the wall beside you. Attracting depends on the mass och each body and distance between them.
Dlovan
Are you really asking if two bodies have to be charged to be influenced by Coulombs Law?
Robert
like charges repel while unlike charges atttact
Raymond
What is specific heat capacity
Destiny Reply
Specific heat capacity is a measure of the amount of energy required to raise the temperature of a substance by one degree Celsius (or Kelvin). It is measured in Joules per kilogram per degree Celsius (J/kg°C).
AI-Robot
specific heat capacity is the amount of energy needed to raise the temperature of a substance by one degree Celsius or kelvin
ROKEEB
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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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