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McGraw Hill Glencoe Geometry, 2012

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McGraw Hill Glencoe Geometry, 2012 View details

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Exercise 52 Page P21

Does either of the equations have an isolated variable in it?

No solution

Practice makes perfect

In this system of equations, at least one of the variables has a coefficient of $1 .$ Therefore, we will approach its solution with the Substitution Method. When solving a system of equations using substitution, there are three steps.

Isolate a variable in one of the equations.
Substitute the expression for that variable into the other equation and solve.
Substitute this solution into one of the equations and solve for the value of the other variable.

Observing the given equations, it looks like it will be simplest to isolate

d

in the second equation.

{6 d + 3 f = 12 2 d = 8 - f (I) (II)

$(II):$ $LHS / 2 = RHS / 2$

{6 d + 3 f = 12 d = 4 - \frac{1}{2} f

Now that we've isolated

d

, we can solve the system by substitution.

{6 d + 3 f = 12 d = 4 - \frac{1}{2} f

$(I):$ $d = 4 - \frac{1}{2} f$

{6 (4 - \frac{1}{2} f) + 3 f = 12 d = 4 - \frac{1}{2} f

$(I):$ Distribute $6$

{24 - 6 (\frac{1}{2}) f + 3 f = 12 d = 4 - \frac{1}{2} f

▼

(I):

Simplify

MoveLeftFacToNumOne

$(I):$ $a \cdot \frac{1}{b} = \frac{a}{b}$

{24 - \frac{6}{2} f + 3 f = 12 d = 4 - \frac{1}{2} f

$(I):$ Calculate quotient

{24 - 3 f + 3 f = 12 d = 4 - \frac{1}{2} f

$(I):$ Add terms

{24 = 12 d = 4 - \frac{1}{2} f

Solving this system of equations resulted in a contradiction;

24

can never be equal to

12 .

Subchapter links

Exercises p.21-P21