Graphical method of finding solutions.

From \(x-y=1\), \(x=y+1\), and substituting in the first equation gives \(9y=29\). Hence \(y=\frac{29}{9}\) and \(x=\frac{38}{9}\).

The governing concept is finding intercepts by setting the other coordinate equal to zero. For the x-intercept, put y = 0 in 8x − 3y = 24. This gives 8x = 24, so x = 3 and the point is (3, 0). For the y-intercept, put x = 0. Then −3y = 24, so y = −8 and the point is (0, −8). Hence option A is correct. The negative sign on the y-intercept is important: the line crosses the y-axis below the origin. Option B changes both signs, while C incorrectly exchanges the coefficients with the intercept values. Option D has the correct x-intercept but an incorrect positive y-intercept.

Multiplying the first equation by (2) gives (10x+4y=18), while the second is (10x+4y=25). Hence the lines are parallel and have no solution.

From \(x+4y=22\), \(x=22-4y\), and substituting in the first equation gives \(y=\frac{73}{16}\). Then \(x=\frac{15}{4}\).

Here \(x-y=-\frac{7}{3}-\frac{2}{3}=-3\). Values of (x) and (y) are read directly from the intersection point.

Subtracting the equations gives \(x=8\), then \(8+5y=26\) gives \(y=\frac{18}{5}\). This is the meeting point of both paths.

Here (\frac{2}{4}=\frac{3}{6}\neq\frac{12}{18}), so the lines are parallel. In exams, check ratios first.

The second equation is (2) times the first, so both lines are the same. If one line overlaps the other, there are infinitely many solutions.

Adding both equations gives (2x=8), so (x=4) and (y=3). On the graph, this is the intersection point.

Both lines have the same slope (2), but different intercepts. Equal slope and different intercepts mean parallel lines.

The vertical line (x=5) and the horizontal line (y=-3) intersect at ((5,-3)). Remember the order ((x,y)).

The second equation is (2) times the first, so the lines are coincident. In coincident lines, every point is a solution.

The governing concept is the graphical classification of a pair of linear equations. When two lines intersect at exactly one point, that point gives one ordered pair satisfying both equations. Therefore the system has one unique solution. A pair with at least one solution is called consistent, and because the solution is unique rather than infinitely many, it is called independent. Hence the correct description is “consistent and independent,” option C. An inconsistent pair has no common point, as with distinct parallel lines. A consistent and dependent pair represents coincident lines and has infinitely many common points. “Always parallel” is not a classification for intersecting lines and directly contradicts the stated graph. Thus the single intersection establishes both consistency and independence.

Here (\frac{5}{10}=\frac{2}{4}\neq\frac{10}{25}), so the lines are parallel. Parallel lines have no common point.

Subtracting the second equation from the first gives (x=4), then (y=1). On the graph, the meeting point is ((4,1)).

Infinite solutions occur when both lines are the same line. For this, all three ratios are equal.

The governing graphical principle is that two distinct points satisfying a linear equation determine its straight-line graph. Test (4, 0): 3(4) + 4(0) = 12 + 0 = 12, so it lies on the line. Test (0, 3): 3(0) + 4(3) = 0 + 12 = 12, so it also lies on the line. These are the x-intercept and y-intercept respectively, and they are distinct points, so joining them gives the required graph. Therefore option A is correct. Option B is false because (4, 0) satisfies the equation, and option C is false for the same reason regarding (0, 3). Option D is false because the coordinates are different, not identical.

The second equation is (3) times the first, so both lines are coincident. In such questions, check the multiplier of the whole equation.

Substituting ((-2,5)) gives (x+y=3) and (2x-y=-9), both true. Test the point in both equations.

Both equations give the same line, so all points on that line are solutions. Note that not all ordered pairs, only points on the line, are solutions.