Pair of Linear Equations in Two Variables: questions with answers

Multiply every term by 2, including the constant. This gives 10x + 14y = 12. Changing only the resulting constant gives a separate parallel line.

The partner 10x + 3y = 12 changes the x-coefficient ratio without changing the y-coefficient ratio. Different ratios give exactly one intersection. A whole-equation multiple gives infinitely many common points.

Both variable coefficients scale by 3. For a separate parallel line, the constant must differ from 36. The nonzero shift gives 15x + 21y = 38.

Let child and adult ticket prices be $p$ and $q$ rupees. The two days give $60p+30q=18000$ and $90p+30q=22500$. Subtracting gives $30p=4500$, so $p=150$. Substitute into the first equation to get $q=300$. Today’s collection is $50\times150+20\times300=13500$ rupees. The question asks for today’s collection; 150 is only the solved child ticket price, 90 is the second day’s child ticket count, and 18000 is the first day’s collection.

Let $x$ and $y$ count children and adults. Solve $x+y=250$ and $120x+300y=48000$ to obtain $x=150$.

Supplementary means the two add to $180$ degrees: $x + y = 180$. 'The larger exceeds the smaller by 24' means the larger MINUS the smaller is 24, so $x - y = 24$. Writing $y - x = 24$ says the smaller exceeds the larger, which would give a negative angle.

One linear equation in two variables draws a whole line, and every point of that line is a solution. Choosing any value of $x$ gives a matching $y$, so there is no shortage of solutions -- which is exactly why a second equation is needed before $x$ and $y$ are pinned down.

Substitute the point: $m(4) + 2 = 22$. The $2$ moves across as $-(2)$, giving $4m = 20$ and $m = 5$. Reading the point the other way round, as $x = 2$ and $y = 4$, gives 9 instead.

The solution of a pair is the point lying on both lines, which is where they cross. Reading the crossing against the numbered axes, it is two units to the right of the $y$-axis and one unit above the $x$-axis, so $x = 2$ and $y = 1$. The pair is written with $x$ first, so the answer is $(2, 1)$; $(1, 2)$ is a different point, one unit right and two up.

The crossing is the only point on both lines. Against the numbered axes it lies three units to the right of the $y$-axis and two units BELOW the $x$-axis, so $x = 3$ and $y = -2$. Exchanging the two would give $(-2, 3)$, which is a point to the left of the $y$-axis and above the $x$-axis -- a different quadrant entirely.

The base lies along the $x$-axis, running from $(-2, 0)$ to $(1, 0)$, so it is $3$ units long. The height is the perpendicular distance from the apex down to that base, which is the apex's $y$-coordinate, $4$. The area is $\tfrac12 \times 3 \times 4 = 6$ square units. Dropping the half would give $12$, and using the apex's $x$-coordinate as the height would give $3$.

Rearrange the second equation as $2x-ky=7$. Equal variable-coefficient ratios give $k=-6/5$. The constant ratios differ, so this value gives no solution.

Matching variable coefficients requires $k=3$. Coincidence also requires $3=6$, which the parameter constraint excludes. Therefore no value works.

Multiplying the first equation throughout by $2$ gives $4x + 6y = 14$, and subtracting the second leaves $y = 6$. Substituting back gives $2x = -11$, so $x = -11/2$ -- a fraction, which is exactly the case a graph cannot be read accurately enough to settle. Forgetting to scale the constant would have given $y = -1$.

Multiply the first equation throughout by $3$: $3x + 9y = 21$. Subtracting the second gives $y = 10$, and substituting back gives $x = -23$. Scaling only the left side -- leaving the constant at $7$ -- would have given $y = -4$ and $x = 19$ instead.