Systems of Linear Equations: Chapter Test

The first number of the pair is the value of $x$ and the second is the value of $y$. Substituting $x = 7$ and $y = 2$ leaves the two sides equal.

$2(7) - 5(2) = 14 - 10 = 4$

The other three give $-31$, $-16$ and $-4$.

The $x$ terms are identical, so subtracting the second equation from the first removes them. Both sides must be subtracted in the same order.

$(5x + 2y) - (5x - 3y) = 4 - 19 \quad\Longrightarrow\quad 5y = -15 \quad\Longrightarrow\quad y = -3$

Then $5x - 6 = 4$ gives $x = 2$, and $(2, -3)$ fits both original equations.

The letter $y$ stands for the quantity $2x - 7$, so every $y$ is replaced by that whole quantity, brackets included.

$3x + 2y = 0 \quad\Longrightarrow\quad 3x + 2(2x - 7) = 0 \quad\Longrightarrow\quad 7x - 14 = 0$

One equation counts tickets and the other counts dollars. Each adult ticket contributes $1$ to the count and $9$ to the takings, and each child ticket contributes $1$ and $5$.

$a + c = 140, \qquad 9a + 5c = 1020$

Solving gives $a = 80$ and $c = 60$, which returns $140$ tickets and $720 + 300 = 1020$ dollars.

Multiplying the second equation by $3$ gives $-3x + 12y = 15$. Adding that to the first removes both letters at once.

$(3x - 12y) + (-3x + 12y) = 9 + 15 \quad\Longrightarrow\quad 0 = 24$

The statement is false, so no pair satisfies both. The coefficients ARE proportional, the first equation being the second scaled by $-3$, but the constants do not follow that same scaling: $-3 \times 5 = -15$, not $9$. Proportional coefficients on their own would have left both answers open.

The $z$ terms are $+z$ and $-z$, so adding the two equations removes $z$. Both sides are added.

$(x + y + z) + (2x + y - z) = 2 + (-3) \quad\Longrightarrow\quad 3x + 2y = -1$

Adding removes a variable exactly when its two coefficients sum to zero, which means they are opposites in sign and equal in size. Only $4$ and $-4$ do that.

$(3x + 4y) + (5x - 4y) = 7 + 9 \quad\Longrightarrow\quad 8x = 16$

The other three leave $3x + 12y = 16$, $8x + 8y = 16$ and $7x + 2y = 16$, each still carrying both letters.

Isolating $x$ in the first equation gives $x = 2y + 3$, and putting that quantity into the second leaves one letter.

$3(2y + 3) + 4y = 19 \quad\Longrightarrow\quad 10y = 10 \quad\Longrightarrow\quad y = 1, \quad x = 5$

Both original equations hold: $5 - 2 = 3$ and $15 + 4 = 19$.

The $y$ coefficients are $3$ and $-5$. The smallest size both reach is $15$, so scale each equation by the other's size.

$5(4x + 3y = 5) \Rightarrow 20x + 15y = 25, \qquad 3(6x - 5y = 17) \Rightarrow 18x - 15y = 51$

Adding gives $38x = 76$, so $x = 2$ and $y = -1$.

With $a$ and $p$ the price per kilogram of apples and pears, the two purchases give $2a + 3p = 23$ and $4a + 5p = 41$. Doubling the first matches the $a$ terms.

$(4a + 6p) - (4a + 5p) = 46 - 41 \quad\Longrightarrow\quad p = 5$

Then $2a = 23 - 15$ gives $a = 4$, and $4(4) + 5(5) = 41$ confirms it.

An equation keeps its solutions when every term on both sides is scaled by the same nonzero number, because that step can be undone by scaling back.

$6x - 9y = 15 \quad\Longrightarrow\quad \frac{6x - 9y}{3} = \frac{15}{3} \quad\Longrightarrow\quad 2x - 3y = 5$

The pair $(4, 1)$ shows the others differ: it satisfies $6x - 9y = 15$, but on the left of the three rejected replacements it gives $5$, $15$ and $6$, against right sides of $15$, $5$ and $12$.

No solution needs the two left sides to be scaled copies while the right sides are not. The $y$ terms fix the scale at $6 \div 3 = 2$, so the $x$ terms must match at that same scale.

$a = 2 \times 2 = 4$

With $a = 4$ the first equation is $4x + 6y = 10$, that is $2x + 3y = 5$, and $5 \ne 7$. Had the constant been $14$ instead of $10$, the same coefficients would have given infinitely many solutions.

Subtracting the two reduced equations gives $6x = 12$, so $x = 2$ and then $y = -1$. Putting both into equation $(1)$ recovers the unknown that was eliminated.

$2 + 2(-1) + z = 4 \quad\Longrightarrow\quad z = 4$

The triple $(2, -1, 4)$ satisfies all three original equations: $4$, $15$ and $-1$.

A letter can be isolated without dividing exactly when its coefficient is $1$ or $-1$. Only the $x$ of the second equation qualifies.

$x - 4y = 9 \quad\Longrightarrow\quad x = 4y + 9$

Every other opening either divides by $3$ or by $4$, or removes nothing at all.

With $c$ the total in dollars after $m$ months, the two plans give $c = 25m + 40$ and $c = 25m + 90$. Setting the totals equal removes every $m$ at once.

$25m + 40 = 25m + 90 \quad\Longrightarrow\quad 40 = 90$

That is false, so the system has no solution. The gap of $50$ dollars is there at the start and nothing in either plan ever closes it.

Adding a multiple of the kept equation to the other one is reversible, so it changes no solutions. Here the second equation plus three times the first gives

$(x - 3y) + 3(2x + y) = 14 + 3(7) \quad\Longrightarrow\quad 7x = 35$

and subtracting three times the first brings the original back. The system $2x + y = 7$ with $7x = 35$ still has the single solution $(5, -3)$, which fits both equations as given.

The coefficients already stand in a fixed ratio: $-6 = -\tfrac{3}{2}(4)$ and $9 = -\tfrac{3}{2}(-6)$. Every solution is shared only if the constant follows that same factor.

$k = -\frac{3}{2}(10) = -15$

Then $-6x + 9y = -15$ is the first equation scaled by $-\tfrac{3}{2}$, so the two describe the same line. Any other $k$ makes the lines parallel and distinct, so proportional coefficients on their own decide nothing.

With $p$ pens and $n$ notebooks, the report says $p + n = 9$ and $3p + 7n = 50$. Substituting $p = 9 - n$ leaves one letter.

$3(9 - n) + 7n = 50 \quad\Longrightarrow\quad 4n = 23 \quad\Longrightarrow\quad n = \frac{23}{4}$

The algebra is sound and the pair is a genuine solution of the system. What it fails is the situation: notebooks are counted in whole numbers, so no purchase matches the report.

Two planes are parallel exactly when one equation's coefficients are a multiple of the other's, and that has been excluded, so every pair of planes meets in a line. Those three lines cannot share a point, since a shared point would solve the system, and they cannot cross each other anywhere else either, so they are distinct and parallel.

$\text{no parallel pair} + \text{no common point} \;\Longrightarrow\; \text{three parallel lines of intersection}$

That is each pair of planes meeting in a line, with those three lines running parallel and never meeting.

Replacing $y$ by the quantity equal to it removes both $x$ terms, so no value of $x$ is fixed and only $c$ is.

$9x - 3(3x - 5) = c \quad\Longrightarrow\quad 9x - 9x + 15 = c \quad\Longrightarrow\quad 15 = c$

At $c = 15$ the statement is true for every $x$, and $9x - 3y = 15$ is just $y = 3x - 5$ rescaled, so every pair on that line is a solution. At any other $c$ the statement is false and there is none.