Graphical method of finding solutions.

The direct answer is A: \(\left(5,6\right)\). The meeting point must satisfy both equations. Subtract \(x+y=11\) from \(3x+y=21\): the \(y\)-terms cancel, giving \(2x=10\), so \(x=5\). Substitute this into \(x+y=11\): \(5+y=11\), hence \(y=6\). Therefore the paths meet at \(\left(5,6\right)\). Check: in the first equation, \(3(5)+6=21\); in the second, \(5+6=11\), so the point works in both. Option A is correct. Option B reverses the coordinates; \((6,5)\) gives 23 in the first equation, not 21. Option C, \((4,7)\), satisfies the second equation but gives 19 in the first. Option D, \((7,4)\), satisfies the second equation but gives 25 in the first. Always substitute a proposed ordered pair into both equations, because the first number is \(x\) and the second is \(y\).

Subtracting the second from the first gives \(y=7\), then \(x+7=11\) gives \(x=4\). In any context, the common point is the solution.

Subtracting the equations gives \(x=6\), then \(6+3y=17\) gives \(y=\frac{11}{3}\). Fraction coordinates can also be graphical solutions.

Small errors can occur while reading fractional or decimal coordinates from a graph. Keep the scale clear and read the point carefully.

A point is always written in \(\left(x,y\right)\) order. Reversing coordinates makes the solution wrong.

The direct answer is A: (6,2). Subtract the second equation from the first: (3x+2y)−(x+2y)=22−10. This gives 2x=12, so x=6. Substitute x=6 into x+2y=10: 6+2y=10, hence 2y=4 and y=2. The lines therefore meet at (6,2). Verify it in both equations: 3(6)+2(2)=18+4=22 and 6+2(2)=10. A is correct. B, (2,6), gives 18 in the first equation and 14 in the second, so it fails both. C, (5,3), gives 21 in the first and 11 in the second, so it fails both. D, (4,4), gives 20 in the first and 12 in the second, so it also fails both. The graphical intersection is the ordered pair satisfying both equations simultaneously. Always substitute the answer back into both equations, because a point satisfying only one line is not the solution of the pair. Memory cue: eliminate one variable, solve the other, substitute back.

To find the \(y\)-intercept, put \(x=0\). Then \(7(0)-2y=14\), so \(-2y=14\) and \(y=-7\). Therefore, the \(y\)-intercept is \(\left(0,-7\right)\). Option B has the wrong sign, while \(\left(2,0\right)\) and \(\left(-2,0\right)\) represent possible \(x\)-intercept points, not the \(y\)-intercept. Exam tip: Set \(x=0\) to find the \(y\)-intercept.

Dividing the first equation by (2) gives (x+3y=9), while the second is (x+3y=12). Hence the lines are parallel and have no solution.

The direct answer is A: \(\left(3.5,2.5\right)\). Convert each coordinate separately. Since \(7\div2=3.5\), we have \(\frac72=3.5\). Similarly, \(5\div2=2.5\), so \(\frac52=2.5\). Thus the decimal point is \(\left(3.5,2.5\right)\). Option A preserves both coordinate positions and converts both fractions correctly. Option B swaps the coordinates, changing the point to a different location. Option C incorrectly reads the fraction as 7.2 and 5.2; a fraction bar means division, not a decimal point. Option D gives 0.7 and 0.5, which would be related to dividing by 10, not by 2. The order of an ordered pair matters: first coordinate is \(x\), second is \(y\). A useful check is to multiply each decimal by 2: \(3.5\times2=7\) and \(2.5\times2=5\).

Putting \(x=-3\) gives \(2\left(-3\right)+y=4\), so \(y=10\). In a vertical line, the value of (x) is already fixed.

Subtracting the second equation from the first gives \(y=5\), then \(x+5=8\) gives \(x=3\). In a real situation, the meeting point is the graphical solution.

Subtracting the equations gives \(3x=15\), so \(x=5\) and \(y=5\). On the graph, this is the intersection point of both lines.

For the pair a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the graphical classification depends on the ratios of corresponding coefficients. Here, a₁/a₂ = 5/1 = 5, b₁/b₂ = 10/2 = 5, and c₁/c₂ = 15/3 = 5. Since all three ratios are equal, the two equations are scalar multiples of one another: the first equation is five times the second. Multiplying an equation by a nonzero constant does not change its graph, so both equations represent exactly the same line. Hence every point on that line is a common solution, giving infinitely many solutions. Distinct parallel lines would have only the first two ratios equal, not all three.

The governing concept is point verification for a linear equation: every ordered pair in a valid value table must make the equation true when substituted. For option A, the first pair gives 5(2) + 11 = 10 + 11 = 21. The second gives 5(4) + 1 = 20 + 1 = 21. Thus both points belong to the line, so A is correct. Option B gives 11 for the first pair and 31 for the second, not 21. Option C gives 5 and 46, and option D gives 26 and 24. Since only option A contains two pairs satisfying 5x + y = 21, it is the correct table for plotting the line. This substitution test is more reliable than guessing from coordinates.

The governing concept is equivalence of linear equations. Two equations represent coincident lines when all three corresponding coefficients, including the constant term, are multiplied by the same non-zero number. Multiplying 2x + 5y = 13 by 2 gives 4x + 10y = 26, exactly the equation in option A. Dividing that equation by 2 returns the original equation, so both equations have every point in common and produce the same graph. Option B has the wrong constant, so it represents a parallel line. Option C keeps the left side but changes the constant, also producing a distinct parallel line. Option D changes the coefficient of y and is not an equivalent equation. Hence A is the only correct choice.

The governing concept is the intercept method. To find the x-intercept, set y = 0 because points on the x-axis have zero y-coordinate. Then 8x − 3(0) = 24, so 8x = 24 and x = 3; the point is (3, 0). To find the y-intercept, set x = 0 because points on the y-axis have zero x-coordinate. This gives −3y = 24, hence y = −8 and the point is (0, −8). Therefore option A is correct. The negative y-intercept is essential. Option B changes both signs, option C confuses coefficients with intercepts, and option D gives the wrong sign for the y-intercept and reverses the requested order.

The governing idea is that the intersection point satisfies both equations simultaneously. Since the first line is y = 3, the y-coordinate of the intersection must be 3. Substitute this value into the second equation: 2x − 5(3) = 1, so 2x − 15 = 1, hence 2x = 16 and x = 8. Therefore the intersection is (8, 3), which is option A. Option B reverses the coordinates. Option C keeps y = 3 but does not satisfy the second equation, since 2(5) − 15 = −5. Options D has the coordinates in an unsuitable order and also fails the fixed y = 3 condition. The horizontal line y = 3 makes the coordinate check especially direct.

The governing graphical criterion is the relationship between slopes and intercepts. Two distinct nonvertical lines with equal slopes have the same direction and therefore are parallel. Because their y-intercepts are different, they are not the same line; distinct parallel lines never meet at any point. Hence the pair has no solution, so option B is correct. A unique solution would require the lines to intersect, which normally occurs when their slopes differ. Infinitely many solutions would occur only if both equations represented the very same line, requiring matching intercepts as well. Perpendicular lines have slopes whose product is −1, but here the product is (−4/3)(−4/3) = 16/9, not −1.

The governing method is point verification: a point lies on a line exactly when substituting its coordinates makes the equation true. For (7, 0), 2(7) + 7(0) = 14, so it lies on the line. For (0, 2), the value is 0 + 14 = 14, so it also lies on the line. For (7/2, 1), the value is 2(7/2) + 7(1) = 7 + 7 = 14, so this point lies on the line as well. For (1, 2), the value is 2(1) + 7(2) = 2 + 14 = 16, not 14. Therefore option D is the only point that does not lie on the line. The conclusion follows from exact substitution rather than visual estimation.

The governing concept is that the intersection coordinates satisfy both equations. Solve the pair by elimination. Multiply the second equation by 2: 10x − 4y = 14. Add this to the first equation, 3x + 4y = 25, to obtain 13x = 39, so x = 3. Substitute x = 3 into 5x − 2y = 7: 15 − 2y = 7, hence −2y = −8 and y = 4. Therefore the y-coordinate is 4, making option C correct. The other choices can result from reading the x-coordinate as the requested value, using only one equation incorrectly, or making an arithmetic error during elimination. Substitution into both original equations confirms the point (3, 4).

The governing concept is that the intersection of two non-parallel lines is the ordered pair satisfying both equations simultaneously. From x − y = 1, rearrange to obtain y = x − 1. Substituting this into 3x + 2y = 18 gives 3x + 2(x − 1) = 18, so 3x + 2x − 2 = 18, 5x = 20, and x = 4. Then y = 4 − 1 = 3. Hence the lines meet at (4, 3), so option A is correct. Verification gives 3(4) + 2(3) = 12 + 6 = 18 and 4 − 3 = 1. The other listed points fail one or both equations, so they cannot represent the common intersection.

The governing concept is the relationship between slopes and the number of intersections. Two non-identical lines with equal slopes have the same direction and therefore are parallel. Since the y-intercepts are different, the lines are not the same line; they are distinct parallel lines. Distinct parallel lines never meet, so their pair of equations has no common ordered pair and hence no solution. Therefore option B is correct. A unique solution would require different slopes so that the lines intersect once. Infinitely many solutions would occur only when both the slope and the intercept were equal, producing coincident lines. Perpendicular lines would require the product of their slopes to be −1, which is not true here.

The governing concept is that the graphical solution is the intersection point, which must satisfy both equations. Add the equations to eliminate y: (3x − y) + (2x + y) = 10 + 15. This gives 5x = 25, so x = 5. Substitute x = 5 into 3x − y = 10: 15 − y = 10, hence y = 5. Therefore the lines intersect at (5, 5), so option A is correct. Verification in the second equation gives 2(5) + 5 = 15, confirming the point. Options B, C, and D do not satisfy both equations simultaneously, even if one may appear close on an inaccurately drawn graph.

The correct answer is A, (6, 3). The intersection point must satisfy both equations at the same time. Test or solve systematically. From x − y = 3, we get y = x − 3. Substituting in the first equation gives 5x + 2(x − 3) = 36, so 7x − 6 = 36, 7x = 42, and x = 6. Then y = 6 − 3 = 3. Therefore the intersection is (6,3). Option A satisfies both: 5(6)+2(3)=36 and 6−3=3. Option B gives 20+12=32, not 36. Option C gives 15+12=27, not 36. Option D gives 25+8=33, not 36. A graphically obtained intersection should always be checked by substitution in both equations.

The governing concept is the graphical condition for a pair of linear equations. Two non-identical straight lines having equal slopes are parallel. Here m₁ = m₂ = −7/4, while the y-intercepts are different, so the lines cannot be the same line; they remain distinct parallel lines. Distinct parallel lines never meet at any point in the Cartesian plane. Since a common solution would be represented by an intersection point, the pair has no common solution. Option B is therefore correct. Option A would describe lines with unequal slopes that intersect once, option C would describe coincident lines with the same slope and the same intercept, and option D is incorrect because perpendicular lines have slopes whose product is −1, not equal slopes.