Graphical method of finding solutions.

The graphical solution of two linear equations is their common point, or the point where the two corresponding lines intersect. We can verify each candidate by substitution. For (3, 5), the first equation gives 2(3) + 5 = 6 + 5 = 11, and the second gives 3 + 5 = 8. Therefore (3, 5) lies on both lines and is their intersection point. Option B is correct. For (5, 3), the first equation gives 13; for (4, 4), it gives 12; and for (2, 6), it gives 10, so those points fail the first equation even though some may satisfy the second. Since a graphical solution must satisfy both equations simultaneously, only option B is valid.

The meeting point of two graphs is the ordered pair that satisfies both linear equations. Test option A: with x = 5 and y = -2, the first equation becomes 5 - 2(-2) = 5 + 4 = 9, and the second becomes 2(5) + (-2) = 10 - 2 = 8. Thus (5, -2) lies on both lines and is their intersection. Option A is correct. The order matters: the first coordinate is x and the second is y. For (-5, 2), the first equation gives -9; for (4, -1), it gives 6; and for (2, 5), it gives -8, so the other options do not satisfy the first equation. A negative y-coordinate is perfectly acceptable and represents a point below the x-axis.

The intersection point must satisfy both equations at the same time. Check option B, (3, 5): in the first equation, 2x - y = 2(3) - 5 = 6 - 5 = 1; in the second equation, x + y = 3 + 5 = 8. Thus (3, 5) is common to both lines and option B is correct. The alternatives can be rejected by substitution: (2, 6) gives 2x - y = -2, (4, 4) gives 4, and (5, 3) gives 7, none of which equals 1 in the first equation. Although these pairs may look plausible because their coordinates add to 8 in some cases, the first equation is also required. Hence the common point is uniquely (3, 5).

A point lies on a line if its coordinates satisfy the equation of that line. Test each candidate by substituting its x- and y-coordinates into 4x − 5y. For (5, 4), the expression is 4(5) − 5(4) = 20 − 20 = 0, so this point satisfies the equation and option B is correct. For (4, 5), the value is 16 − 25 = −9. For (2, 5), it is 8 − 25 = −17, and for (5, 2), it is 20 − 10 = 10. Since none of these values is zero, the other three points do not lie on the line. Direct substitution is the decisive verification method.

To create points for plotting the line x + 3y = 12, substitute each given x-value and solve for y. When x = 3, we get 3 + 3y = 12, so 3y = 9 and y = 3; this gives the point (3, 3). When x = 6, we get 6 + 3y = 12, so 3y = 6 and y = 2; this gives (6, 2). Therefore option A is correct. Each ordered pair must use the selected x-value as its first coordinate and the calculated y-value as its second coordinate. The other options either use an incorrect y-value or fail the equation when substituted. For example, (3, 6) gives 21 rather than 12, so it cannot lie on the line.

A value table for a line is valid only when every listed ordered pair satisfies the equation. Rearranging 4x + y = 16 gives y = 16 − 4x. For x = 1, y = 16 − 4(1) = 12, so (1, 12) lies on the line. For x = 3, y = 16 − 4(3) = 4, so (3, 4) also lies on it. Therefore option A is correct. In option B, substituting (1, 4) gives 8 rather than 16, and substituting (3, 12) gives 24. In option C, the expressions give 4 and 32; option D also fails substitution. Checking each pair directly prevents exchanging or guessing coordinate values.

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.

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 governing concept is the definition of a y-intercept. It is the point where a graph crosses the y-axis, and every point on the y-axis has x-coordinate 0. Therefore, substitute x = 0 into 2x − 5y = 10: 2(0) − 5y = 10, so −5y = 10 and y = −2. Hence the y-intercept is (0, −2), making option A correct. Option B has y = 0, so it represents an x-axis intercept rather than a y-intercept. Option C results from ignoring the negative sign while dividing by −5. Option D is the x-intercept because putting y = 0 gives 2x = 10 and x = 5. Thus both the coordinate definition and direct substitution confirm option A.

The governing rule is that an x-intercept is found by setting y = 0, while a y-intercept is found by setting x = 0. For the x-intercept, 3x − 4(0) = 12, so 3x = 12 and x = 4; the point is therefore (4, 0). For the y-intercept, 3(0) − 4y = 12, so −4y = 12 and y = −3; the point is (0, −3). Thus option A is correct. Option B reverses both signs. Option C incorrectly treats the coefficients as the intercept coordinates. Option D contains incorrect signs and also presents the two intercepts in the reverse order requested. Substitution verifies both coordinates: 3(4) − 4(0) = 12 and 3(0) − 4(−3) = 12.

The governing coordinate-plane fact is that every point on the x-axis has y-coordinate equal to zero. The x-coordinate may be any real number, so it is represented by a parameter such as a. Consequently, if two lines meet on the x-axis, their common point must have the form (a, 0), making option B correct. Option A describes a general point on the y-axis because its x-coordinate is zero. Option C has equal coordinates and lies on the line y = x, not necessarily on the x-axis. Option D also has a generally nonzero y-coordinate, so it cannot lie on the x-axis unless a is zero, which is not the required general form.

The governing concept is slope-intercept form, y = mx + c, where m is the slope and c is the y-intercept. In option A, both equations have m = 2, so their slopes are equal, while their y-intercepts are 5 and −3, which are different. These are therefore distinct parallel lines and have no common point. In option B, the slopes are 2 and −2, so the slopes are not equal. In option C, the slopes are 1 and 2, again unequal. In option D, both the slope and intercept are identical, so the equations represent the same line, not two distinct lines. Hence option A is the only pair satisfying both stated conditions.

The governing concept is simultaneous satisfaction: an intersection point must lie on both graphs. The equation y = −2 represents a horizontal line, so any common point must have y-coordinate −2. Substitute this value into 4x + 3y = 10: 4x + 3(−2) = 10, giving 4x − 6 = 10, then 4x = 16 and x = 4. Therefore the intersection is (4, −2), so option A is correct. Verification gives 4(4) + 3(−2) = 16 − 6 = 10. Option D has the correct y-coordinate but gives 20 − 6 = 14, not 10. Options B and C do not lie on the horizontal line y = −2.

The answer is D, (4, 2). To test whether a point lies on a line, substitute its x- and y-coordinates into the equation. If the left side becomes exactly 24, the point lies on the line; if it gives another value, the point does not lie on it. For A, 3(8)+8(0)=24+0=24, so (8,0) lies on the line. For B, 3(0)+8(3)=0+24=24, so (0,3) also lies on it. For C, 3(8/3)+8(2)=8+16=24, so this point is also valid. For D, 3(4)+8(2)=12+16=28, not 24. Therefore D is the only point that does not lie on the line. A useful check is to remember that a point is verified by substitution, not by looking at its coordinates separately.

The intersection must satisfy both equations. The equation y = -4 represents a horizontal line, so the y-coordinate of every point on it is fixed at -4. Substitute y = -4 into 5x + 2y = 17: 5x + 2(-4) = 17, so 5x - 8 = 17. Hence 5x = 25 and x = 5. The intersection point is therefore (5, -4), making option A correct. Option B reverses the coordinate order. Option C keeps the correct y-value but gives an x-value that does not satisfy the second equation, while D fails both coordinate requirements. Verification gives 5(5) + 2(-4) = 25 - 8 = 17.