AlgebraDifficulty 10-60432 tagged problems

Systems of Equations

Systems of Equations is the habit of solving multiple linked equations by elimination, substitution, symmetry, or strategic combination. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases} \implies x = \frac{c_1 b_2 - c_2 b_1}{a_1 b_2 - a_2 b_1}, \quad y = \frac{a_1 c_2 - a_2 c_1}{a_1 b_2 - a_2 b_1}$$

Several equations sharing unknowns can be combined — by elimination or substitution — to pin down each unknown's value.

Why It Works

1
To eliminate $y$, scale the equations so the $y$-coefficients match: multiply the first by $b_2$ and the second by $b_1$.
2
Subtract: $(a_1 b_2 - a_2 b_1)x = c_1 b_2 - c_2 b_1$, which isolates $x$ (provided $a_1 b_2 - a_2 b_1 \ne 0$).
3
By the same elimination of $x$, $(a_1 b_2 - a_2 b_1)y = a_1 c_2 - a_2 c_1$.
4
Dividing by the common factor $a_1 b_2 - a_2 b_1$ gives Cramer's-rule formulas; in practice just add/subtract scaled equations to cancel a variable.

Worked Examples

Example 1

Solve $\begin{cases} 2x + 3y = 13 \\ x - y = 1 \end{cases}$.

1
From the second equation, $x = y + 1$ (substitution).
2
Substitute into the first: $2(y + 1) + 3y = 13$, i.e. $5y + 2 = 13$, so $5y = 11$ and $y = \frac{11}{5}$.
3
Back-substitute: $x = y + 1 = \frac{11}{5} + 1 = \frac{16}{5}$. (Check: $2\cdot\frac{16}{5} + 3\cdot\frac{11}{5} = \frac{32 + 33}{5} = \frac{65}{5} = 13$.)

Answer:

$x = \dfrac{16}{5},\; y = \dfrac{11}{5}$.

Warm-up

Solve $\begin{cases} 3x + 2y = 16 \\ 5x - 2y = 8 \end{cases}$.

1
The $y$-coefficients are already opposite ($+2$ and $-2$), so add the equations to eliminate $y$: $(3x + 5x) = 16 + 8$, i.e. $8x = 24$.
2
Solve: $x = 3$.
3
Back-substitute into $3x + 2y = 16$: $9 + 2y = 16$, so $2y = 7$ and $y = \frac{7}{2}$. (Check: $5(3) - 2(\frac72) = 15 - 7 = 8$.)

Answer:

$x = 3,\ y = \dfrac{7}{2}$.

Contest level

Solve $\begin{cases} x + y + z = 6 \\ x + 2y + 3z = 14 \\ x + 4y + 9z = 36 \end{cases}$.

1
Subtract equation $1$ from equation $2$: $(y + 2z) = 8$. Subtract equation $2$ from equation $3$: $(2y + 6z) = 22$, i.e. $y + 3z = 11$.
2
Subtract these two reduced equations: $(y + 3z) - (y + 2z) = 11 - 8$, giving $z = 3$.
3
Back-substitute: $y + 2(3) = 8 \Rightarrow y = 2$, then $x + 2 + 3 = 6 \Rightarrow x = 1$. (Check eq. $3$: $1 + 8 + 27 = 36$.)

Answer:

$x = 1,\ y = 2,\ z = 3$.

Olympiad / Challenge

Find all real pairs $(x, y)$ with $x + y = 5$ and $x^2 + y^2 = 13$.

1
This is symmetric, so introduce $s = x + y = 5$ and $p = xy$. Use $x^2 + y^2 = s^2 - 2p$: $13 = 25 - 2p$, so $p = 6$.
2
Then $x$ and $y$ are the roots of $t^2 - st + p = 0$, i.e. $t^2 - 5t + 6 = 0$.
3
Factor: $(t - 2)(t - 3) = 0$, so $\{x, y\} = \{2, 3\}$. (Check: $2 + 3 = 5$, $4 + 9 = 13$.)

Answer:

$(x, y) = (2, 3)$ or $(3, 2)$.

Going Deeper

Generalization: a linear system $A\mathbf{x} = \mathbf{b}$ has a unique solution iff $\det A \ne 0$ (Cramer's rule); the determinant $a_1 b_2 - a_2 b_1$ in the statement is exactly the $2\times 2$ case. Nonlinear systems often yield to the same elimination or to symmetric substitution $s = x+y,\ p = xy$.
Where it appears: AMC word problems (mixtures, rates, coins), AIME setups disguised as geometry or sequences, and olympiad symmetric systems that collapse via Vieta once you switch to $s$ and $p$.
Pitfall: when $\det = a_1 b_2 - a_2 b_1 = 0$ the lines are parallel — the system has either NO solution or INFINITELY many; do not blindly divide by the determinant. And a symmetric substitution can manufacture spurious pairs: discard any $(s, p)$ giving a negative discriminant if you need real solutions.

Spot the Signal

Look for problems where the key step is solving multiple linked equations by elimination, substitution, symmetry, or strategic combination.
You can describe the hard part as solving multiple linked equations by elimination, substitution, symmetry, or strategic combination, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the expression or equation that calls for systems of equations, then rewrite the givens around it.
Name the relation that makes Systems of Equations legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Systems of Equations just because the surface looks familiar; verify the required condition first.
Applying Systems of Equations because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let systems of equations reveal the algebraic structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is solving multiple linked equations by elimination, substitution, symmetry, or strategic combination.

Back to techniques

AlgebraDifficulty 10-60432 tagged problems

Systems of Equations

Practice this technique Match my level

The Key Result

$$\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases} \implies x = \frac{c_1 b_2 - c_2 b_1}{a_1 b_2 - a_2 b_1}, \quad y = \frac{a_1 c_2 - a_2 c_1}{a_1 b_2 - a_2 b_1}$$

Several equations sharing unknowns can be combined — by elimination or substitution — to pin down each unknown's value.

Why It Works

1
To eliminate $y$, scale the equations so the $y$-coefficients match: multiply the first by $b_2$ and the second by $b_1$.
2
Subtract: $(a_1 b_2 - a_2 b_1)x = c_1 b_2 - c_2 b_1$, which isolates $x$ (provided $a_1 b_2 - a_2 b_1 \ne 0$).
3
By the same elimination of $x$, $(a_1 b_2 - a_2 b_1)y = a_1 c_2 - a_2 c_1$.
4
Dividing by the common factor $a_1 b_2 - a_2 b_1$ gives Cramer's-rule formulas; in practice just add/subtract scaled equations to cancel a variable.

Worked Examples

Example 1

Solve $\begin{cases} 2x + 3y = 13 \\ x - y = 1 \end{cases}$.

1
From the second equation, $x = y + 1$ (substitution).
2
Substitute into the first: $2(y + 1) + 3y = 13$, i.e. $5y + 2 = 13$, so $5y = 11$ and $y = \frac{11}{5}$.
3
Back-substitute: $x = y + 1 = \frac{11}{5} + 1 = \frac{16}{5}$. (Check: $2\cdot\frac{16}{5} + 3\cdot\frac{11}{5} = \frac{32 + 33}{5} = \frac{65}{5} = 13$.)

Answer:

$x = \dfrac{16}{5},\; y = \dfrac{11}{5}$.

Warm-up

Solve $\begin{cases} 3x + 2y = 16 \\ 5x - 2y = 8 \end{cases}$.

1
The $y$-coefficients are already opposite ($+2$ and $-2$), so add the equations to eliminate $y$: $(3x + 5x) = 16 + 8$, i.e. $8x = 24$.
2
Solve: $x = 3$.
3
Back-substitute into $3x + 2y = 16$: $9 + 2y = 16$, so $2y = 7$ and $y = \frac{7}{2}$. (Check: $5(3) - 2(\frac72) = 15 - 7 = 8$.)

Answer:

$x = 3,\ y = \dfrac{7}{2}$.

Contest level

Solve $\begin{cases} x + y + z = 6 \\ x + 2y + 3z = 14 \\ x + 4y + 9z = 36 \end{cases}$.

1
Subtract equation $1$ from equation $2$: $(y + 2z) = 8$. Subtract equation $2$ from equation $3$: $(2y + 6z) = 22$, i.e. $y + 3z = 11$.
2
Subtract these two reduced equations: $(y + 3z) - (y + 2z) = 11 - 8$, giving $z = 3$.
3
Back-substitute: $y + 2(3) = 8 \Rightarrow y = 2$, then $x + 2 + 3 = 6 \Rightarrow x = 1$. (Check eq. $3$: $1 + 8 + 27 = 36$.)

Answer:

$x = 1,\ y = 2,\ z = 3$.

Olympiad / Challenge

Find all real pairs $(x, y)$ with $x + y = 5$ and $x^2 + y^2 = 13$.

1
This is symmetric, so introduce $s = x + y = 5$ and $p = xy$. Use $x^2 + y^2 = s^2 - 2p$: $13 = 25 - 2p$, so $p = 6$.
2
Then $x$ and $y$ are the roots of $t^2 - st + p = 0$, i.e. $t^2 - 5t + 6 = 0$.
3
Factor: $(t - 2)(t - 3) = 0$, so $\{x, y\} = \{2, 3\}$. (Check: $2 + 3 = 5$, $4 + 9 = 13$.)

Answer:

$(x, y) = (2, 3)$ or $(3, 2)$.

Going Deeper

Generalization: a linear system $A\mathbf{x} = \mathbf{b}$ has a unique solution iff $\det A \ne 0$ (Cramer's rule); the determinant $a_1 b_2 - a_2 b_1$ in the statement is exactly the $2\times 2$ case. Nonlinear systems often yield to the same elimination or to symmetric substitution $s = x+y,\ p = xy$.
Where it appears: AMC word problems (mixtures, rates, coins), AIME setups disguised as geometry or sequences, and olympiad symmetric systems that collapse via Vieta once you switch to $s$ and $p$.
Pitfall: when $\det = a_1 b_2 - a_2 b_1 = 0$ the lines are parallel — the system has either NO solution or INFINITELY many; do not blindly divide by the determinant. And a symmetric substitution can manufacture spurious pairs: discard any $(s, p)$ giving a negative discriminant if you need real solutions.

Spot the Signal

Look for problems where the key step is solving multiple linked equations by elimination, substitution, symmetry, or strategic combination.
You can describe the hard part as solving multiple linked equations by elimination, substitution, symmetry, or strategic combination, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the expression or equation that calls for systems of equations, then rewrite the givens around it.
Name the relation that makes Systems of Equations legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Systems of Equations just because the surface looks familiar; verify the required condition first.
Applying Systems of Equations because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let systems of equations reveal the algebraic structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is solving multiple linked equations by elimination, substitution, symmetry, or strategic combination.