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Pearson Algebra 2 Common Core, 2011

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Pearson Algebra 2 Common Core, 2011 View details

2. Standard Form of a Quadratic Function

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Exercise 60 Page 208

In this system of equations, at least one of the variables has a coefficient of $1 .$ Moreover, it is isolated. 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.

For this exercise,

y

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = \frac{2}{3} x - 3 - x + 3 y = 18 \frac{2}{3} (I) (II)

$(II):$ $y = \frac{2}{3} x - 3$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = \frac{2}{3} x - 3 - x + 3 (\frac{2}{3} x - 3) = 18 (I) (II)

$(II):$ Distribute $3$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = \frac{2}{3} x - 3 - x + 2 x - 9 = 18 \frac{2}{3} (I) (II)

$(II):$ Add terms

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = \frac{2}{3} x - 3 x - 9 = 18 \frac{2}{3} (I) (II)

$(II):$ $LHS + 9 = RHS + 9$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = \frac{2}{3} x - 3 x = 27 \frac{2}{3} (I) (II)

Great! Now, to find the value of

y,

we need to substitute

x = 27

into either one of the equations in the given system. Let's use the first equation.

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = \frac{2}{3} x - 3 x = 27 \frac{2}{3} (I) (II)

$(I):$ $x = 27$

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = \frac{2}{3} (27) - 3 x = 27 \frac{2}{3} (I) (II)

MoveRightFacToNum

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

⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ y = \frac{5 4}{3} - 3 x = 27 \frac{2}{3} (I) (II)

$(I):$ Calculate quotient

{y = 18 - 3 x = 27 (I) (II)

Subtract term

{y = 15 x = 27 (I) (II)

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

(27, 15) .

Subchapter links

Lesson Check p.206

Practice and Problem-Solving Exercises p.206-208

Standardized Test Prep p.208

Mixed Review p.208