Download Systems of Equations: Eg 2x + 3y = 1 x − 3y = 5 To solve this system

Systems of Equations:
E.g.
2x + 3y = 1
x − 3y = 5
To solve this system of two equations with two unknowns what
we are looking for are values of (x, y) that would solve both equations.
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For example, the ordered pair ( , 0) solves the first equation, but does
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not solve the second one, so it is not a solution to the system. The
ordered pair (2, −1) solves both the top and bottom equation, so it is
a solution.
Geometrically, a solution to an equation is a point on the graph of the
equation, so a solution to the first equation is a point on the graph
of the first equation, and a solution to the second equation is a point
on the second equation. Therefore, a solution to the system of equation is a point of intersection of the two graphs. To solve systems
of equations, geometrically, is to find point(s) of intersection of the
graphs.
If two lines intersect at one point the two equations has one solution.
We have an independent system.
If two lines are parallel we have no solution. The system of equations
is inconsistent.
If two lines are the same line we have infinitely many solutions. The
system is dependent.
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Algebraically, we may use the method of substitution or the method
of addition (sometimes called the method of elimination) to solve
the system.
Method of Substitution:
2x + 3y = 1
x − 3y = 5
Solve for one of the variable in terms of the other using one of the
equations, then substitute for that variable in the other equation.
Using the second equation:
x = 3y + 5
Substitute the expression for x (in terms of y) in the first equation:
2(3y + 5) + 3y = 1
6y + 10 + 3y = 1
9y + 10 = 1
9y = −9
y = −1
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Once we found the value of y, we may substitute it into any of the two
equations to solve for x:
2x + 3(−1) = 1
2x − 3 = 1
2x = 4
x=2
The solution is (2, −1), notice that the solution is an ordered pair.
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Method of Addition:
4x − 3y = 6
3x − 2y = 5
We add the two equations with one of the variables having coefficients
of the same magnitude but opposite sign (for example, one maybe 4x
and the other −4x) so we may eliminate one of the variables. If the
two equations do not have any variables with coefficients of opposite
sign, we multiply by the LCM of the coefficient of one of the variables.
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Multiply first equation by −3, multiply second equation by 4:
−3[4x − 3y = 6]
4[3x − 2y = 5]
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We get a new system of equations:
−12x + 9y = −18
12x − 8y = 20
We now add the two equations and this eliminates the x variable:
y=2
Knowing the value of y we may once again substitute the value into
one of the two equations to solve for the value of x:
4x − 3(2) = 6
4x − 6 = 6
4x = 12
x=3
The solution is (3, 2)
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