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 Amsco Integrated Algebra I: Using Addition to Solve a System of Linear Equations
This is a free lesson from our course in Amsco's Integrated Algebra
 
   
This lesson explains the basics and process of Solving a system of linear equations using Addition. Here you’ll learn about using the addition method where the two equations in the system are added to create a new equation with only one variable. To achieve this, the other variable must cancel out i.e. you’ll need to first perform operations on each equation until one term has an equal and opposite coefficient/sign for the corresponding term in the other equation.
(More text below video...)
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(Continued from above) To solve systems of equations using the addition method, follow the below given steps:
• rearrange the given equations so that the variables are in the same order and the constant is on the other side.
• multiply one or both equations (as required) by an integer such that one term has equal and opposite coefficients/sign in the two equations.
• add the equations to have a single equation with one variable.
• solve the above resulting equation with the variable.
• replace this variable by its value in either of the given equations and solve for second variable.
• check the solution by substituting value of both the variables in the given equations - it should satisfy both equations.
For example: Given- solve the system of equations
5a + b = 13 [A]
4a - 3b = 18 [B]
Multiply both sides of equation [A] by 3 to obtain an equivalent equation, i.e. 15a + 3b = 39
Add the corresponding members of equations and eliminate b,
15a + 3b = 39
4a - 3b = 18
The resulting equation is: 19a = 57 and solving this gives, a = 3.
Replace a by its value in either of the given equations and solve for b, it yields b=-2
Check: Substituting 3 for a and -2 for b in the given equations, satisfies both the equations i.e. a= 3 and b= -2 is the final answer.
The video above explains; more details on Using Addition to solve a system of linear Equations, with the help of several examples and their solutions
 
   

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