🧮 Algebra Systems Linear Nonlinear (12 Focused Problems)
Recommended: 30–40 minutes. No calculator.
Problems
1.
Tags: Linear System · Easy · source: Original (AMC-style)
If $x + y = 7$ and $x - y = 3$, what is $x$?
A) $2$
B) $3$
C) $4$
D) $5$
E) $6$
Answer & Solution
Answer: D
Adding the equations: $(x+y) + (x-y) = 7 + 3$, so $2x = 10$ and $x = 5$.
2.
Tags: Linear System · Easy · source: Original (AMC-style)
If $2x + 3y = 13$ and $3x + 2y = 12$, what is $x + y$?
A) $3$
B) $4$
C) $5$
D) $6$
E) $7$
Answer & Solution
Answer: C
Adding the equations: $(2x+3y) + (3x+2y) = 13 + 12$, so $5x + 5y = 25$. Dividing by 5: $x + y = 5$.
3.
Tags: Linear System · Easy · source: Original (AMC-style)
If $x + 2y = 8$ and $2x - y = 1$, what is $y$?
A) $1$
B) $2$
C) $3$
D) $4$
E) $5$
Answer & Solution
Answer: C
From the second equation: $x = \frac{y+1}{2}$. Substituting into the first: $\frac{y+1}{2} + 2y = 8$, so $y+1 + 4y = 16$, giving $5y = 15$ and $y = 3$.
4.
Tags: Linear System · Easy · source: Original (AMC-style)
If $3x + 4y = 20$ and $2x + 3y = 14$, what is $x$?
A) $2$
B) $3$
C) $4$
D) $5$
E) $6$
Answer & Solution
Answer: C
Subtracting the second equation from the first: $(3x+4y) - (2x+3y) = 20 - 14$, so $x + y = 6$. From the second equation: $2x + 3y = 14$. Substituting: $2x + 3(6-x) = 14$, so $2x + 18 - 3x = 14$, giving $-x = -4$ and $x = 4$.
5.
Tags: Nonlinear System · Medium · source: Original (AMC-style)
If $x^2 + y^2 = 25$ and $x + y = 7$, what is $xy$?
A) $6$
B) $8$
C) $10$
D) $12$
E) $14$
Answer & Solution
Answer: D
We have $(x+y)^2 = x^2 + 2xy + y^2 = 49$. Since $x^2 + y^2 = 25$, we get $25 + 2xy = 49$, so $2xy = 24$ and $xy = 12$.
6.
Tags: Nonlinear System · Medium · source: Original (AMC-style)
If $x^2 - y^2 = 15$ and $x - y = 3$, what is $x + y$?
A) $3$
B) $5$
C) $7$
D) $9$
E) $11$
Answer & Solution
Answer: B
We have $x^2 - y^2 = (x-y)(x+y) = 15$. Since $x - y = 3$, we get $3(x+y) = 15$, so $x + y = 5$.
7.
Tags: Linear System · Medium · source: Original (AMC-style)
If $x + y + z = 6$, $x + 2y + 3z = 14$, and $2x + y + z = 8$, what is $z$?
A) $1$
B) $2$
C) $3$
D) $4$
E) $5$
Answer & Solution
Answer: B
Subtracting the first equation from the second: $(x+2y+3z) - (x+y+z) = 14 - 6$, so $y + 2z = 8$. Subtracting the first equation from the third: $(2x+y+z) - (x+y+z) = 8 - 6$, so $x = 2$. From the first equation: $2 + y + z = 6$, so $y + z = 4$. From $y + 2z = 8$ and $y + z = 4$: $z = 4$.
8.
Tags: Nonlinear System · Medium · source: Original (AMC-style)
If $xy = 12$ and $x^2 + y^2 = 25$, what is $(x+y)^2$?
A) $37$
B) $49$
C) $61$
D) $73$
E) $85$
Answer & Solution
Answer: B
We have $(x+y)^2 = x^2 + 2xy + y^2 = 25 + 2(12) = 25 + 24 = 49$.
9.
Tags: Complex System · Hard · source: Original (AMC-style)
If $x + y + z = 9$, $x^2 + y^2 + z^2 = 29$, and $xyz = 6$, what is $x^3 + y^3 + z^3$?
A) $27$
B) $45$
C) $63$
D) $81$
E) $99$
Answer & Solution
Answer: C
We have $(x+y+z)^2 = x^2 + y^2 + z^2 + 2(xy+yz+zx) = 81$. Since $x^2 + y^2 + z^2 = 29$, we get $xy + yz + zx = 26$. Now $x^3 + y^3 + z^3 = (x+y+z)^3 - 3(x+y+z)(xy+yz+zx) + 3xyz = 9^3 - 3(9)(26) + 3(6) = 729 - 702 + 18 = 45$.
10.
Tags: Quadratic System · Hard · source: Original (AMC-style)
If $x^2 + y^2 = 13$ and $xy = 6$, what are the possible values of $x + y$?
A) $\pm 5$
B) $\pm 4$
C) $\pm 3$
D) $\pm 2$
E) $\pm 1$
Answer & Solution
Answer: A
We have $(x+y)^2 = x^2 + 2xy + y^2 = 13 + 2(6) = 25$. So $x + y = \pm 5$.
11.
Tags: Advanced Nonlinear · Hard · source: Original (AMC-style)
If $x^2 + y^2 = 10$ and $x^3 + y^3 = 28$, what is $x + y$?
A) $2$
B) $3$
C) $4$
D) $5$
E) $6$
Answer & Solution
Answer: C
Let $s = x + y$ and $p = xy$. We have $s^2 - 2p = 10$ and $s^3 - 3sp = 28$. From the first: $p = \frac{s^2 - 10}{2}$. Substituting into the second: $s^3 - 3s \cdot \frac{s^2 - 10}{2} = 28$, so $2s^3 - 3s(s^2 - 10) = 56$. This gives $2s^3 - 3s^3 + 30s = 56$, so $-s^3 + 30s = 56$, or $s^3 - 30s + 56 = 0$. Trying $s = 4$: $64 - 120 + 56 = 0$ ✓.
12.
Tags: System with Parameters · Hard · source: Original (AMC-style)
If $x + y = a$ and $x^2 + y^2 = b$, what is $x^4 + y^4$ in terms of $a$ and $b$?
A) $2b^2 - a^2b$
B) $2b^2 - a^4$
C) $b^2 - 2a^2b + a^4$
D) $2b^2 - 4a^2b + a^4$
E) $b^2 - a^4$
Answer & Solution
Answer: D
We have $xy = \frac{a^2 - b}{2}$ from $(x+y)^2 = x^2 + 2xy + y^2$. Now $x^4 + y^4 = (x^2)^2 + (y^2)^2 = (x^2 + y^2)^2 - 2x^2y^2 = b^2 - 2(xy)^2 = b^2 - 2\left(\frac{a^2 - b}{2}\right)^2 = b^2 - \frac{(a^2 - b)^2}{2} = b^2 - \frac{a^4 - 2a^2b + b^2}{2} = \frac{2b^2 - a^4 + 2a^2b - b^2}{2} = \frac{b^2 - a^4 + 2a^2b}{2}$.
Answer Key
| # | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Ans | D | C | C | C | D | B | B | B | C | A | C | D |