Graphical method of finding solutions.

A pair of linear equations has a solution at every common point of their graphs. The equation y=2 is a horizontal line containing points whose y-coordinate is 2, while y=9 is another horizontal line at height 9. Because 2 and 9 are different, no point can lie on both lines: a common point would need to have y=2 and y=9 simultaneously. The lines are distinct and parallel, so the system has zero solutions. Hence option B is correct. Infinitely many solutions would occur if both equations represented the same line, one solution would require intersecting lines, and two solutions cannot occur for two distinct linear lines.

In an ordered pair (x, y), the first coordinate represents x and the second coordinate represents y. Therefore, in (3, 10), the value of y is 10. Option 3 represents the value of x, not y. In an exam, remember that the coordinates of an intersection point are written in the order (x, y).

In an ordered pair \((x,y)\), the first coordinate represents \(x\) and the second represents \(y\). Therefore, in \((12,1)\), \(x=12\). Exam tip: Always read the coordinates in their given order.

The second equation is (2) times the first, so the lines are coincident. Coincident lines have infinitely many solutions.

Multiplying the first equation by (2) gives (2x-2y=12), which differs from (2x-2y=15). Therefore, the lines are parallel.

Substituting x=4 in x+4y=8 gives 4+4y=8. Thus, 4y=4 and y=1, so the point is (4, 1). Option A also has x=4, but 4+4×2=12, which does not equal 8. Exam tip: verify a point on a line by substituting both its coordinates into the original equation.

Substituting \(y=1\) in \(4x+y=17\) gives \(4x+1=17\), so \(4x=16\) and \(x=4\). Hence, the ordered pair is \((4,1)\). Option B reverses the order of the coordinates, while options C and D do not satisfy the equation. Exam tip: verify a point by substituting both coordinates into the original equation.

The point of intersection must satisfy both equations. Subtracting the equations gives \(x-y=0\), so \(x=y\). Substituting this in \(x+2y=6\) gives \(3x=6\), hence \(x=y=2\). Therefore, the lines intersect at \((2,2)\). Exam tip: verify a candidate point in both equations; \((4,1)\) satisfies the first equation but not the second.

Subtracting the first equation from the second gives \(x=4\). Substituting this into \(x+3y=10\) gives \(4+3y=10\), so \(y=2\). Therefore, the lines intersect at \((4,2)\). The distractor \((2,4)\) satisfies the first equation but gives \(2(2)+3(4)=16\neq14\), so it is not the common solution. In an exam, verify an intersection point in both equations.

A clear scale helps plot points at the correct places. A wrong scale can cause mistakes in reading the intersection point.

The coordinates of a point are always read in the order \((x,y)\): the x-coordinate first and the y-coordinate second. Therefore, reading \((5,3)\) as \((3,5)\) means reading the coordinates in reverse order. Naming the axes or choosing the correct scale does not describe this particular error. Exam tip: record the horizontal-coordinate value first, followed by the vertical-coordinate value.

The governing test is direct substitution: a point lies on a line if its coordinates make the equation true. For the origin (0,0), x+y=0+0=0, so the equation is satisfied and the line must pass through that point. Therefore option C is correct. Checking the distractors confirms the result: (1,1) gives 2, (0,1) gives 1, and (1,0) gives 1; none of these equals 0. In fact, x+y=0 can also be written as y=−x, a line through the origin with slope −1. The phrase “must pass” asks for a guaranteed point, and the origin is guaranteed by the zero constant term.

For the point \((3,2)\), substitute \(x=3\) and \(y=2\): \(2x-3y=2(3)-3(2)=6-6=0\). Hence, the point lies on the line. For example, substituting \((2,1)\) gives \(4-3=1\), so it does not lie on the line. In an exam, the quickest method is to substitute each point into the equation.

Exactly one solution is obtained when both lines intersect at one point. That point is the common solution of both equations.

Distinct parallel lines never intersect, so they have no common point and the pair of equations has no solution. Coincident lines have infinitely many solutions, while intersecting lines have one unique solution. Exam tip: lines with the same slope but different intercepts represent no solution.

A pair of linear equations is represented by two straight lines. A solution is a point that lies on both lines at the same time. If the two equations produce exactly the same line, every point on that line belongs to both graphs. Consequently, every one of those points satisfies both equations.

Since a straight line contains infinitely many points, the pair has infinitely many common solutions. Therefore, option B, “when the lines are the same line,” is correct. Distinct parallel lines never meet and give no solution, while ordinary intersecting lines meet at one point and give one solution. Merely cutting the axes does not determine the number of common solutions.

Subtracting the second equation from the first gives \(y=4\). Substituting this into \(x+y=8\) gives \(x=4\), so the graphs intersect at \((4,4)\). Options B and D satisfy the second equation but not the first. Exam tip: Verify a graphical solution by substituting both coordinates into both equations.

The point where two lines intersect is the common solution of both linear equations. Therefore, the point (4, 6) gives x = 4 and y = 6. Option B reverses the order of the coordinates. In an exam, remember to write an ordered pair in the order (x, y).

From (5+y=13), (y=8), so the point is ( (5,8) ). While drawing a graph, write the point as ( (x,y) ).

Substituting \(x=3\) gives \(2(3)+y=14\), so \(6+y=14\) and therefore \(y=8\). Option A results from not completing the subtraction correctly after finding \(2x=6\). For graphing, the corresponding ordered pair is \((3,8)\).