AMC 10 · 2006 · #12
Grade 8 arithmeticThe lines x=41y+a and y=41x+b intersect at the point (1,2). What is a+b?
Pick an answer.
AMC 10 2006 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 are given by $x=\frac{1}{4}y+a$ and $y=\frac{1}{4}x+b$, and they cross each other at the point $(1,2)$. Find the value of $a+b$.
Givens: The first line is $x=\frac{1}{4}y+a$; The second line is $y=\frac{1}{4}x+b$; The two lines intersect at the point $(1,2)$, so $x=1$ and $y=2$ there; Answer choices: (A) $0$, (B) $\frac{3}{4}$, (C) $1$, (D) $2$, (E) $\frac{9}{4}$
Unknowns: The constant $a$; The constant $b$; Their sum $a+b$
Understand
Restated: Two lines are given by $x=\frac{1}{4}y+a$ and $y=\frac{1}{4}x+b$, and they cross each other at the point $(1,2)$. Find the value of $a+b$.
Givens: The first line is $x=\frac{1}{4}y+a$; The second line is $y=\frac{1}{4}x+b$; The two lines intersect at the point $(1,2)$, so $x=1$ and $y=2$ there; Answer choices: (A) $0$, (B) $\frac{3}{4}$, (C) $1$, (D) $2$, (E) $\frac{9}{4}$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #15 Organize Information in More Ways, #3 Eliminate Possibilities
The point $(1,2)$ is where the lines meet, which means it sits on each line — so its coordinates must satisfy both equations. Tool #4 (Introduce a Variable) is really about setting up and solving equations for an unknown: here $a$ and $b$ are the unknowns, and substituting $x=1,\,y=2$ turns each two-variable line into a one-line equation that hands over $a$ (from the first) and $b$ (from the second). Add them for $a+b$. Tool #15 (Organize Information in More Ways) offers a slicker route — add the two original equations before substituting so $a+b$ appears together in one step. Tool #3 (Eliminate Possibilities) then confirms the result against the five listed choices.
Execute — Answer: E
8.EE.C.8 Step 1 Use that the point lies on both lines
- Saying the lines intersect at $(1,2)$ means this single point is on the first line and on the second line at the same time.
- A point is on a line exactly when its coordinates make that line's equation true.
- So $x=1$ together with $y=2$ must satisfy both equations.
- That lets us drop $x$ and $y$ in and read off the leftover constants.
💡 An intersection point is the one place both lines agree, so it obeys both rules at once.
6.EE.B.7 Step 2 Plug the point into the first line to get a
- The first line is $x=\frac{1}{4}y+a$.
- Put in $x=1$ and $y=2$: $1=\frac{1}{4}\cdot 2+a=\frac{1}{2}+a$.
- This is now a simple equation with only $a$ unknown.
- Subtract $\frac{1}{2}$ from both sides to get $a=1-\frac{1}{2}=\frac{1}{2}$.
💡 With the point substituted, the only thing left to balance the equation is $a$ itself.
6.EE.B.7 Step 3 Plug the point into the second line to get b
- The second line is $y=\frac{1}{4}x+b$.
- Put in $x=1$ and $y=2$: $2=\frac{1}{4}\cdot 1+b=\frac{1}{4}+b$.
- Subtract $\frac{1}{4}$ from both sides: $b=2-\frac{1}{4}=\frac{8}{4}-\frac{1}{4}=\frac{7}{4}$.
💡 The same substitution trick empties the second equation of everything but $b$.
5.NF.A.1 Step 4 Add a and b
- Now add the two constants.
- $a+b=\frac{1}{2}+\frac{7}{4}$.
- Rewrite $\frac{1}{2}$ as $\frac{2}{4}$ so the denominators match: $\frac{2}{4}+\frac{7}{4}=\frac{9}{4}$.
- So $a+b=\frac{9}{4}$, which is choice (E).
💡 To add fractions you first make the bottoms match, then just add the tops.
8.EE.C.8 Saying the lines intersect at $(1,2)$ means this single point is on the first li 6.EE.B.7 The first line is $x=\frac{1}{4}y+a$. Put in $x=1$ and $y=2$: $1=\frac{1}{4}\cdo 6.EE.B.7 The second line is $y=\frac{1}{4}x+b$. Put in $x=1$ and $y=2$: $2=\frac{1}{4}\cd 5.NF.A.1 Now add the two constants. $a+b=\frac{1}{2}+\frac{7}{4}$. Rewrite $\frac{1}{2}$ Review
Reasonableness: Check the constants back in the lines. First line: $\frac{1}{4}(2)+\frac{1}{2}=\frac{1}{2}+\frac{1}{2}=1=x$. Second line: $\frac{1}{4}(1)+\frac{7}{4}=\frac{1}{4}+\frac{7}{4}=\frac{8}{4}=2=y$. Both equations hold at $(1,2)$, so $a=\frac{1}{2}$ and $b=\frac{7}{4}$ are right, and $a+b=\frac{9}{4}$. The sum is a little more than $2$, which fits it being the largest choice, (E).
Alternative: Add the two original equations before substituting. $x+y=\frac{1}{4}y+a+\frac{1}{4}x+b=\frac{1}{4}(x+y)+(a+b)$, so $\frac{3}{4}(x+y)=a+b$. At the point $(1,2)$, $x+y=3$, giving $a+b=\frac{3}{4}\cdot 3=\frac{9}{4}$ in one move — choice (E). This shortcut finds $a+b$ directly without ever computing $a$ and $b$ separately.
CCSS standards used (min grade 8)
8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Recognizing that the intersection point $(1,2)$ lies on both lines and so must satisfy both equations at once.)6.EE.B.7Solve real-world and mathematical problems by writing and solving equations of the form $x + p = q$ (Substituting the point to get $1=\frac{1}{2}+a$ and $2=\frac{1}{4}+b$, then solving each for the constant.)5.NF.A.1Add and subtract fractions with unlike denominators (Combining $a+b=\frac{1}{2}+\frac{7}{4}=\frac{2}{4}+\frac{7}{4}=\frac{9}{4}$ by matching denominators.)
⭐ An intersection point sits on both lines, so plug its coordinates into each equation — everything else drops out and the unknown constants fall right into your hands.
⭐ An intersection point sits on both lines, so plug its coordinates into each equation — everything else drops out and the unknown constants fall right into your hands.
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