AMC 10 · 2014 · #21
Grade 8 arithmeticPositive integers a and b are such that the graphs of y=ax+5 and y=3x+b intersect the x-axis at the same point. What is the sum of all possible x-coordinates of these points of intersection?
Pick an answer.
AMC 10 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Two lines, $y=ax+5$ and $y=3x+b$ with positive integers $a$ and $b$, cross the $x$-axis at the same point. Find the sum of every possible $x$-coordinate where that shared crossing can happen.
Givens: The line $y=ax+5$ and the line $y=3x+b$; $a$ and $b$ are positive integers; Both lines cross the $x$-axis at the same point; Answer choices: (A) $-20$, (B) $-18$, (C) $-15$, (D) $-12$, (E) $-8$
Unknowns: Every possible $x$-coordinate of the shared $x$-axis crossing; The sum of all those $x$-coordinates
Understand
Restated: Two lines, $y=ax+5$ and $y=3x+b$ with positive integers $a$ and $b$, cross the $x$-axis at the same point. Find the sum of every possible $x$-coordinate where that shared crossing can happen.
Givens: The line $y=ax+5$ and the line $y=3x+b$; $a$ and $b$ are positive integers; Both lines cross the $x$-axis at the same point; Answer choices: (A) $-20$, (B) $-18$, (C) $-15$, (D) $-12$, (E) $-8$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #2 Make a Systematic List, #7 Identify Subproblems
The phrase "cross the $x$-axis at the same point" is a geometry sentence, so Tool #4 (Introduce a Variable) converts it into algebra: write each line's $x$-intercept in terms of $a$ and $b$, then set them equal. That collapses the whole condition into one equation, $ab=15$. Tool #2 (Make a Systematic List) then finds every positive-integer pair $(a,b)$ with $ab=15$, guaranteeing we catch all possible crossing points. Tool #7 (Identify Subproblems) keeps the work in clean stages — intercept, factor pairs, coordinates, sum — instead of one tangled calculation.
Execute — Answer: E
8.EE.C.7 Step 1 Find each line's x-axis crossing
- A point on the $x$-axis has height $y=0$.
- Put $y=0$ into each line and solve for $x$.
- For $y=ax+5$: $0=ax+5$ gives $x=-\dfrac{5}{a}$.
- For $y=3x+b$: $0=3x+b$ gives $x=-\dfrac{b}{3}$.
- Each formula is the exact spot where that line meets the $x$-axis.
💡 Crossing the $x$-axis just means the height is zero, so set $y=0$ and solve for $x$.
8.EE.C.7 Step 2 Set the two crossings equal
- The problem says both lines hit the $x$-axis at the same point, so their $x$-coordinates match: $-\dfrac{5}{a}=-\dfrac{b}{3}$.
- Cancel the minus signs and cross-multiply: $5\cdot 3 = a\cdot b$, so $ab=15$.
- The whole shared-crossing condition is really just the single fact that $a$ times $b$ equals $15$.
💡 Two equal fractions can be cross-multiplied, turning the geometry into one tidy equation $ab=15$.
4.OA.B.4 Step 3 List the positive factor pairs of 15
- Since $a$ and $b$ are positive integers with $ab=15$, list every way to write $15$ as (positive whole number) times (positive whole number).
- The divisors of $15$ are $1,3,5,15$, giving the pairs $(a,b)=(1,15),(3,5),(5,3),(15,1)$.
- These four pairs are all the lines the problem allows.
💡 Every valid pair is a factor pair of $15$, so just list the divisors of $15$.
5.NF.B.3 Step 4 Turn each pair into its x-coordinate
- The crossing point is $x=-\dfrac{5}{a}$, so substitute each $a$.
- $a=1\Rightarrow x=-5$; $a=3\Rightarrow x=-\dfrac{5}{3}$; $a=5\Rightarrow x=-1$; $a=15\Rightarrow x=-\dfrac{1}{3}$.
- All four are different, so there are four distinct crossing points.
💡 A fraction like $-\dfrac{5}{a}$ is just $-5$ split into $a$ equal parts.
7.NS.A.1 Step 5 Add the x-coordinates
- Add the four values: $-5-\dfrac{5}{3}-1-\dfrac{1}{3}$.
- Group the whole numbers, $-5-1=-6$, and the thirds, $-\dfrac{5}{3}-\dfrac{1}{3}=-\dfrac{6}{3}=-2$.
- Together $-6+(-2)=-8$.
- So the sum of all possible $x$-coordinates is $-8$, which is choice (E).
💡 Adding negatives just piles up more of the same, so combine like pieces and total them.
8.EE.C.7 A point on the $x$-axis has height $y=0$. Put $y=0$ into each line and solve for 8.EE.C.7 The problem says both lines hit the $x$-axis at the same point, so their $x$-coo 4.OA.B.4 Since $a$ and $b$ are positive integers with $ab=15$, list every way to write $1 5.NF.B.3 The crossing point is $x=-\dfrac{5}{a}$, so substitute each $a$. $a=1\Rightarrow 7.NS.A.1 Add the four values: $-5-\dfrac{5}{3}-1-\dfrac{1}{3}$. Group the whole numbers, Review
Reasonableness: Each crossing sits at a negative $x$ between $-5$ and $-\tfrac{1}{3}$, so four of them must add to something negative and no smaller than $4\times(-5)=-20$; $-8$ lands comfortably inside that range. A common slip is forgetting the reversed pairs $(5,3)$ and $(15,1)$, but the four distinct values $-5,-\tfrac{5}{3},-1,-\tfrac{1}{3}$ are exactly the factor pairs of $15$, and they sum to $-8$, matching (E).
Alternative: Factor the sum instead of adding term by term: $-\dfrac{5}{1}-\dfrac{5}{3}-\dfrac{5}{5}-\dfrac{5}{15}=-5\left(1+\tfrac{1}{3}+\tfrac{1}{5}+\tfrac{1}{15}\right)=-5\cdot\dfrac{15+5+3+1}{15}=-5\cdot\dfrac{24}{15}=-8$. Same total, confirming (E).
CCSS standards used (min grade 8)
8.EE.C.7Solve linear equations in one variable (Setting $y=0$ to solve each line for its $x$-intercept, and clearing fractions in $-\dfrac{5}{a}=-\dfrac{b}{3}$ to reach $ab=15$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Listing every positive factor pair of $15$: $(1,15),(3,5),(5,3),(15,1)$.)5.NF.B.3Interpret a fraction as division of the numerator by the denominator (Reading $x=-\dfrac{5}{a}$ as $-5$ divided by $a$ to get $-5,-\dfrac{5}{3},-1,-\dfrac{1}{3}$.)7.NS.A.1Apply and extend understanding of addition and subtraction to rational numbers (Adding the negative integers and fractions to reach the total $-8$.)
⭐ Turn "same x-axis crossing" into the equation $ab=15$, list its factor pairs, then add up each crossing point.
⭐ Turn "same x-axis crossing" into the equation $ab=15$, list its factor pairs, then add up each crossing point.
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