Systems of linear equations

Answer: $\text{infinitely many solutions}$

The second equation's left side is $-1$ times the first's. So is its right side $(-1 \times 4 = -4)\mathclose{:}$ it is the SAME line, and every point on it is a solution.

Common mistakes, and the answer each one gives:

• Saw two different equations and concluded the lines never meet → $\text{no solution}$
• Assumed two lines always cross exactly once → $\text{exactly one solution}$

Answer: $-2$

The first equation already gives $y\mathclose{,}$ so put it straight into the second: $x + 5\left(4 x + 10\right) = 8\mathclose{.}$ Expand: $21 x + 50 = 8\mathclose{,}$ so $21 x = -42$ and $x = -2\mathclose{.}$ (Then the first equation gives $y = 2\mathclose{.})$

Common mistakes, and the answer each one gives:

• Gave the value of $y$ instead of $x$ → $2$
• Multiplied only the first term of the bracket by $5$ → $- \frac{2}{21}$

Answer: $-6$

Multiply the first by $3\mathclose{,}$ so the $y$ terms become $3 y$ and $- 3 y\mathclose{,}$ then add: $19 x = -114\mathclose{,}$ so $x = -6\mathclose{.}$ Put that into the first equation: $y - 36 = -34\mathclose{,}$ so $y = 2\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the value of $y$ instead of $x$ → $2$
• Multiplied only the left-hand sides when scaling the equations → $- \frac{46}{19}$

Answer: $y = - 2 x + 3$

The second equation is $-1$ times the first, so only one condition remains: $4 x + 2 y = 6\mathclose{,}$ which gives $y = - 2 x + 3$ — one solution for every $x\mathclose{.}$

Common mistakes, and the answer each one gives:

• Lost the sign of the $x$ term moving it across → $y = 2 x + 3$

Answer: $6$

No solution means parallel lines: the left sides must be multiples of each other. The $x$ coefficients are in the ratio $-2\mathclose{,}$ so $k = -2 \times (-3) = 6\mathclose{.}$ (The right sides are then $-2$ against $3\mathclose{,}$ so the lines are parallel and distinct.)

Common mistakes, and the answer each one gives:

• Took the coefficient of $y$ from the first equation unchanged → $-3$
• Divided by the factor instead of multiplying → $\dfrac{3}{2}$

Answer: $\text{no solution}$

The second equation's left side is $2$ times the first's. Its right side is not $(2 \times 9 = 18\mathclose{,}$ not $23)\mathclose{:}$ the lines are parallel and never meet.

Common mistakes, and the answer each one gives:

• Compared the left-hand sides only → $\text{infinitely many solutions}$
• Assumed two lines always cross exactly once → $\text{exactly one solution}$

Answer: $-6$

The first equation already gives $y\mathclose{,}$ so put it straight into the second: $5 x + \left(- 4 x - 24\right) = -30\mathclose{.}$ Expand: $x - 24 = -30\mathclose{,}$ so $x = -6\mathclose{.}$ (Then the first equation gives $y = 0\mathclose{.})$

Common mistakes, and the answer each one gives:

• Gave the value of $y$ instead of $x$ → $0$

Answer: $-3$

The $y$ terms are already $y$ and $- y\mathclose{,}$ so add the two equations: $6 x = 6\mathclose{,}$ so $x = 1\mathclose{.}$ Put that into the first equation: $y + 3 = 0\mathclose{,}$ so $y = -3\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the value of $x$ instead of $y$ → $1$

Answer: $y = - \frac{2 x}{3} - \frac{8}{3}$

The second equation is $2$ times the first, so only one condition remains: $- 2 x - 3 y = 8\mathclose{,}$ which gives $y = - \frac{2 x}{3} - \frac{8}{3}$ — one solution for every $x\mathclose{.}$

Common mistakes, and the answer each one gives:

• Lost the sign of the $x$ term moving it across → $y = \frac{2 x}{3} - \frac{8}{3}$

Answer: $-6$

No solution means parallel lines: the left sides must be multiples of each other. The $x$ coefficients are in the ratio $3\mathclose{,}$ so $k = 3 \times (-2) = -6\mathclose{.}$ (The right sides are then $-18$ against $-13\mathclose{,}$ so the lines are parallel and distinct.)

Common mistakes, and the answer each one gives:

• Took the coefficient of $y$ from the first equation unchanged → $-2$
• Divided by the factor instead of multiplying → $-\dfrac{2}{3}$

Answer: $\text{no solution}$

The second equation's left side is $-1$ times the first's. Its right side is not $(-1 \times 6 = -6\mathclose{,}$ not $-10)\mathclose{:}$ the lines are parallel and never meet.

Common mistakes, and the answer each one gives:

• Compared the left-hand sides only → $\text{infinitely many solutions}$
• Assumed two lines always cross exactly once → $\text{exactly one solution}$

Answer: $6$

The first equation already gives $y\mathclose{,}$ so put it straight into the second: $4 x + 4\left(x - 5\right) = 28\mathclose{.}$ Expand: $8 x - 20 = 28\mathclose{,}$ so $8 x = 48$ and $x = 6\mathclose{.}$ (Then the first equation gives $y = 1\mathclose{.})$

Common mistakes, and the answer each one gives:

• Gave the value of $y$ instead of $x$ → $1$
• Multiplied only the first term of the bracket by $4$ → $\frac{33}{8}$

Answer: $-1$

Multiply the first by $6\mathclose{,}$ so the $y$ terms become $6 y$ and $- 6 y\mathclose{,}$ then add: $14 x = -70\mathclose{,}$ so $x = -5\mathclose{.}$ Put that into the first equation: $y - 10 = -11\mathclose{,}$ so $y = -1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the value of $x$ instead of $y$ → $-5$
• Multiplied only the left-hand sides when scaling the equations → $- \frac{62}{7}$

Answer: $y = \frac{3 x}{5} - \frac{8}{5}$

The second equation is $3$ times the first, so only one condition remains: $3 x - 5 y = 8\mathclose{,}$ which gives $y = \frac{3 x}{5} - \frac{8}{5}$ — one solution for every $x\mathclose{.}$

Common mistakes, and the answer each one gives:

• Lost the sign of the $x$ term moving it across → $y = - \frac{3 x}{5} - \frac{8}{5}$

Answer: $4$

No solution means parallel lines: the left sides must be multiples of each other. The $x$ coefficients are in the ratio $-2\mathclose{,}$ so $k = -2 \times (-2) = 4\mathclose{.}$ (The right sides are then $-10$ against $-14\mathclose{,}$ so the lines are parallel and distinct.)

Common mistakes, and the answer each one gives:

• Took the coefficient of $y$ from the first equation unchanged → $-2$
• Divided by the factor instead of multiplying → $1$

Answer: $\text{infinitely many solutions}$

The second equation's left side is $-1$ times the first's. So is its right side $(-1 \times (-2) = 2)\mathclose{:}$ it is the SAME line, and every point on it is a solution.

Common mistakes, and the answer each one gives:

• Saw two different equations and concluded the lines never meet → $\text{no solution}$
• Assumed two lines always cross exactly once → $\text{exactly one solution}$

Answer: $5$

The first equation already gives $y\mathclose{,}$ so put it straight into the second: $x + 4\left(4 - x\right) = 1\mathclose{.}$ Expand: $16 - 3 x = 1\mathclose{,}$ so $- 3 x = -15$ and $x = 5\mathclose{.}$ (Then the first equation gives $y = -1\mathclose{.})$

Common mistakes, and the answer each one gives:

• Gave the value of $y$ instead of $x$ → $-1$

Answer: $4$

Multiply the second by $3\mathclose{,}$ so the $y$ terms become $3 y$ and $- 3 y\mathclose{,}$ then add: $19 x = 38\mathclose{,}$ so $x = 2\mathclose{.}$ Put that into the first equation: $3 y + 8 = 20\mathclose{,}$ so $3 y = 12$ and $y = 4\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the value of $x$ instead of $y$ → $2$
• Multiplied only the left-hand sides when scaling the equations → $\frac{92}{19}$

Answer: $y = \frac{3 x}{4} - \frac{7}{4}$

The second equation is $-1$ times the first, so only one condition remains: $3 x - 4 y = 7\mathclose{,}$ which gives $y = \frac{3 x}{4} - \frac{7}{4}$ — one solution for every $x\mathclose{.}$

Common mistakes, and the answer each one gives:

• Lost the sign of the $x$ term moving it across → $y = - \frac{3 x}{4} - \frac{7}{4}$

Answer: $-3$

No solution means parallel lines: the left sides must be multiples of each other. The $x$ coefficients are in the ratio $3\mathclose{,}$ so $k = 3 \times (-1) = -3\mathclose{.}$ (The right sides are then $-27$ against $-24\mathclose{,}$ so the lines are parallel and distinct.)

Common mistakes, and the answer each one gives:

• Took the coefficient of $y$ from the first equation unchanged → $-1$
• Divided by the factor instead of multiplying → $-\dfrac{1}{3}$

Answer: $\text{no solution}$

The second equation's left side is $3$ times the first's. Its right side is not $(3 \times 8 = 24\mathclose{,}$ not $20)\mathclose{:}$ the lines are parallel and never meet.

Common mistakes, and the answer each one gives:

• Compared the left-hand sides only → $\text{infinitely many solutions}$
• Assumed two lines always cross exactly once → $\text{exactly one solution}$

Answer: $3$

The first equation already gives $y\mathclose{,}$ so put it straight into the second: $3 x + 3\left(8 - 2 x\right) = 15\mathclose{.}$ Expand: $24 - 3 x = 15\mathclose{,}$ so $- 3 x = -9$ and $x = 3\mathclose{.}$ (Then the first equation gives $y = 2\mathclose{.})$

Common mistakes, and the answer each one gives:

• Gave the value of $y$ instead of $x$ → $2$
• Multiplied only the first term of the bracket by $3$ → $-9$

Answer: $1$

Multiply the second by $3\mathclose{,}$ so the $y$ terms become $6 y$ and $- 6 y\mathclose{,}$ then add: $8 x = -56\mathclose{,}$ so $x = -7\mathclose{.}$ Put that into the first equation: $6 y - 14 = -8\mathclose{,}$ so $6 y = 6$ and $y = 1\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the value of $x$ instead of $y$ → $-7$
• Multiplied only the left-hand sides when scaling the equations → $- \frac{1}{3}$

Answer: $y = \frac{3 x}{5} + \frac{6}{5}$

The second equation is $3$ times the first, so only one condition remains: $- 3 x + 5 y = 6\mathclose{,}$ which gives $y = \frac{3 x}{5} + \frac{6}{5}$ — one solution for every $x\mathclose{.}$

Common mistakes, and the answer each one gives:

• Lost the sign of the $x$ term moving it across → $y = - \frac{3 x}{5} + \frac{6}{5}$

Answer: $4$

No solution means parallel lines: the left sides must be multiples of each other. The $x$ coefficients are in the ratio $-1\mathclose{,}$ so $k = -1 \times (-4) = 4\mathclose{.}$ (The right sides are then $-1$ against $1\mathclose{,}$ so the lines are parallel and distinct.)

Common mistakes, and the answer each one gives:

• Took the coefficient of $y$ from the first equation unchanged → $-4$

Answer: $\text{infinitely many solutions}$

The second equation's left side is $2$ times the first's. So is its right side $(2 \times (-6) = -12)\mathclose{:}$ it is the SAME line, and every point on it is a solution.

Common mistakes, and the answer each one gives:

• Saw two different equations and concluded the lines never meet → $\text{no solution}$
• Assumed two lines always cross exactly once → $\text{exactly one solution}$

Answer: $-3$

The first equation already gives $y\mathclose{,}$ so put it straight into the second: $2 x + 5\left(- 4 x - 7\right) = 19\mathclose{.}$ Expand: $- 18 x - 35 = 19\mathclose{,}$ so $- 18 x = 54$ and $x = -3\mathclose{.}$ (Then the first equation gives $y = 5\mathclose{.})$

Common mistakes, and the answer each one gives:

• Gave the value of $y$ instead of $x$ → $5$
• Multiplied only the first term of the bracket by $5$ → $- \frac{13}{9}$

Answer: $-3$

Multiply the first by $2$ and the second by $3\mathclose{,}$ so the $y$ terms become $6 y$ and $- 6 y\mathclose{,}$ then add: $26 x = -78\mathclose{,}$ so $x = -3\mathclose{.}$ Put that into the first equation: $3 y - 12 = 3\mathclose{,}$ so $3 y = 15$ and $y = 5\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the value of $y$ instead of $x$ → $5$
• Multiplied only the left-hand sides when scaling the equations → $- \frac{25}{26}$

Answer: $y = \frac{5 x}{3} - \frac{2}{3}$

The second equation is $2$ times the first, so only one condition remains: $- 5 x + 3 y = -2\mathclose{,}$ which gives $y = \frac{5 x}{3} - \frac{2}{3}$ — one solution for every $x\mathclose{.}$

Common mistakes, and the answer each one gives:

• Lost the sign of the $x$ term moving it across → $y = - \frac{5 x}{3} - \frac{2}{3}$

Answer: $-3$

No solution means parallel lines: the left sides must be multiples of each other. The $x$ coefficients are in the ratio $-1\mathclose{,}$ so $k = -1 \times 3 = -3\mathclose{.}$ (The right sides are then $-8$ against $-12\mathclose{,}$ so the lines are parallel and distinct.)

Common mistakes, and the answer each one gives:

• Took the coefficient of $y$ from the first equation unchanged → $3$