Exercise 9 Page 209

When solving a system of equations using the Substitution Method, 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.

For this exercise,

x

is already isolated in one equation, so we can skip straight to solving!

{x = 2 y + 7 3 x - 2 y = 3 (I) (II)

Substitute

$(II):$ $x = 2 y + 7$

{x = 2 y + 7 3 (2 y + 7) - 2 y = 3

Distr

$(II):$ Distribute $3$

{x = 2 y + 7 3 (2 y) + 3 (7) - 2 y = 3

▼

(II):

Simplify left-hand side

Multiply

$(II):$ Multiply

{x = 2 y + 7 6 y + 21 - 2 y = 3

CommutativePropAdd

$(II):$ Commutative Property of Addition

{x = 2 y + 7 6 y - 2 y + 21 = 3

SubTerm

$(II):$ Subtract term

{x = 2 y + 7 4 y + 21 = 3

SubEqn

$(II):$ $LHS - 21 = RHS - 21$

{x = 2 y + 7 4 y + 21 - 21 = 3 - 21

SubTerm

$(II):$ Subtract term

{x = 2 y + 7 4 y = - 18

Now, we can use the Division Property of Equality to simplify the equation even further.

{x = 2 y + 7 4 y = - 18

DivEqn

$(II):$ $LHS / 4 = RHS / 4$

{x = 2 y + 7 \frac{4 y}{4} = \frac{- 1 8}{4}

▼

Simplify left-hand side

MovePartNumRight

$(II):$ $\frac{a \cdot b}{c} = \frac{a}{c} \cdot b$

{x = 2 y + 7 \frac{4}{4} y = \frac{- 1 8}{4}

CalcQuot

$(II):$ Calculate quotient

{x = 2 y + 7 1 y = \frac{- 1 8}{4}

IdPropMult

$(II):$ Identity Property of Multiplication

{x = 2 y + 7 y = \frac{- 1 8}{4}

▼

Simplify right-hand side

MoveNegNumToFrac

$(II):$ Put minus sign in front of fraction

{x = 2 y + 7 y = - \frac{1 8}{4}

ReduceFrac

$(II):$ $\frac{a}{b} = \frac{a / 2}{b / 2}$

{x = 2 y + 7 y = - \frac{9}{2}

Great! Now, to find the value of

x,

we need to substitute

y = - \frac{9}{2}

into the first equation.

{x = 2 y + 7 y = - \frac{9}{2}

Substitute

$(I):$ $y = - \frac{9}{2}$

{x = 2 (- \frac{9}{2}) + 7 y = - \frac{9}{2}

MoveNegFracToNum

$(I):$ Put minus sign in numerator

{x = 2 (\frac{- 9}{2}) + 7 y = - \frac{9}{2}

DenomMultFracToNumber

$(I):$ $2 \cdot \frac{a}{2} = a$

{x = - 9 + 7 y = - \frac{9}{2}

AddTerms

$(I):$ Add terms

{x = - 2 y = - \frac{9}{2}

The solution, or point of intersection, to this system of equations is the point

(- 2, - \frac{9}{2}) .

Checking Our Answer

To check our answer, we will substitute our solution into both equations. If doing so results in true statements, then our solution is correct.

{x = 2 y + 7 3 x - 2 y = 3 (I) (II)

$(I), (II):$ $x = - 2$ , $y = - \frac{9}{2}$

{- 2 = ? 2 (- \frac{9}{2}) + 7 3 (- 2) - 2 (- \frac{9}{2}) = ? 3

$(I), (II):$ $a (- b) = - a \cdot b$

{- 2 = ? - 2 (\frac{9}{2}) + 7 - 3 (2) - 2 (- \frac{9}{2}) = ? 3

MultNegNegOnePar

$(II):$ $- a (- b) = a \cdot b$

{- 2 = ? - 2 (\frac{9}{2}) + 7 - 3 (2) + 2 (\frac{9}{2}) = ? 3

$(I), (II):$ $2 \cdot \frac{a}{2} = a$

{- 2 = ? - 9 + 7 - 3 (2) + 9 = ? 3

Multiply

$(I):$ Multiply

{- 2 = ? - 9 + 7 - 6 + 9 = ? 3

$(I), (II):$ Add terms

{- 2 = - 2 ✓ 3 = 3 ✓

Because both equations are true statements, we know that our solution is correct.