2. Solving Linear Systems by Substitution
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x+y=5 & (I) -6y-6y=30 & (II)
(II):Subtract terms
(II):.LHS /(-12).=.RHS /(-12).
Now that we have isolated the y-variable, let's substitute it into the first equation to solve for the x-variable.
(I):y= -2.5
(I):Subtract terms
The system has one solution, (7.5,-2.5).
(II):x= 7-y
(II):Distribute 5
(II):Add terms
(II):LHS-35=RHS-35
(II):.LHS /(-3).=.RHS /(-3).
Now that the y-variable is solved, let's substitute the value into the first equation and solve for the x-variable.
The system has one solution, (3,12).
Now that we have isolated the y-variable, let's substitute it into the second equation to solve for the x-variable.
(II):y= 5-3x
(II):Distribute 2
(II):Subtract terms
Since 10 is not equal to 12 it is a false statement and there are no solutions.
(I):x= -15-7y
(I):Distribute 2
(I):Add terms
(I):LHS+30=RHS+30
(I):.LHS /(-9).=.RHS /(-9).
Let's substitute the y value into the second equation to solve for x.
(II):y= -2
(II):Multiply
(II):Add terms
We have found that there is one solution, (-1, -2).
3x+5y=17 & (I) -6x-10y=-34 & (II)
(I):LHS * 2=RHS* 2
(I):Distribute 2
(I):Multiply
Now that we have the opposite coefficients we can add the first equation to the second equation.
(II):Add I
(II):Add terms
Since 0 is equal to 0 it is a true statement and therefor the system has infinitely many solutions.