Question

applications

1. laura is thinking of a number such that the sum of the number and five times two more than the number is 26 more than four times the number. determine the number laura is thinking of.

1. as if #2 wasnt confusing enough, laura is now trying to come up with a number where three less than 8 times the number is equal to half of 16 times the number after it was increased by 1. she cant seem to find a number that works. explain why.

1. when finding the intersection of two lines from both algebra i and geometry, you first \set the linear equations equal\ to each other. find the intersection point of the two lines whose equations are shown below. be sure to find both the x and y coordinates.

( y = 5x + 1 ) and ( y = 2x - 11 )

reasoning

1. explain why you cannot find the intersection points of the two lines shown below. give both an algebraic reason and a graphical reason.

( y = 4x + 1 ) and ( y = 4x + 10 )

common core algebra ii, unit #1 - essential algebra concepts - lesson #2
emathinstruction, red hook, ny 12571, © 2015

Explanation:

Problem 2

Step1: Define variable for the number

Let the number be $x$.

Step2: Translate to equation

$x + 5(x+2) = 4x + 26$

Step3: Expand left side

$x + 5x + 10 = 4x + 26$

Step4: Simplify left side

$6x + 10 = 4x + 26$

Step5: Isolate $x$ terms

$6x - 4x = 26 - 10$

Step6: Solve for $x$

$2x = 16 \implies x = \frac{16}{2} = 8$

Step1: Define variable for the number

Let the number be $x$.

Step2: Translate to equation

$8x - 3 = \frac{1}{2}(16(x+1))$

Step3: Simplify right side

$8x - 3 = 8(x+1)$

Step4: Expand right side

$8x - 3 = 8x + 8$

Step5: Isolate $x$ terms

$8x - 8x = 8 + 3$

Step6: Analyze result

$0 = 11$ (false statement)

Step1: Set equations equal

$5x + 1 = 2x - 11$

Step2: Isolate $x$ terms

$5x - 2x = -11 - 1$

Step3: Solve for $x$

$3x = -12 \implies x = \frac{-12}{3} = -4$

Step4: Find $y$ with $x=-4$

$y = 5(-4) + 1 = -20 + 1 = -19$

Answer:

$8$

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Problem 3