Graphical method of finding solutions.

Direct answer: option A, (3,2). The graphical solution of two linear equations is the common point satisfying both equations. Test the coordinates x=3 and y=2 in the first equation: 2x+3y=2(3)+3(2)=6+6=12. Test them in the second: x+y=3+2=5. Both equations are true, so their graphs meet at (3,2). Option A works. Option B, (2,3), gives 2(2)+3(3)=13, not 12, so it fails the first equation. Option C, (4,1), gives 2(4)+3(1)=11, so it fails the first equation. Option D, (1,4), gives 14 in the first equation and also does not give 5 in the second. Thus only A is the intersection. Memory cue: a graph intersection must pass both substitution checks.

In the graphical method, the intersection point of the two lines is the solution of the pair of equations. Adding the equations gives 4x = 12, so x = 3. Substituting this in x + y = 5 gives y = 2. Therefore, the intersection point is (3, 2). The nearby option (2, 3) is incorrect because 3(2) − 3 = 3, not 7. Exam tip: Verify a coordinate pair in both equations before selecting the answer.

Distinct parallel lines never intersect, so there is no common point satisfying both equations. Hence, the pair has no solution. Exam tip: distinct parallel lines represent an inconsistent pair of equations.

The governing concept is comparing two linear equations after reducing them to a common standard form. Divide every term of the first equation 4x + 2y = 10 by 2. This gives 2x + y = 5, which is exactly the second equation. Therefore both equations describe the same set of points and represent one line drawn twice. Such lines are coincident and have infinitely many common solutions, so option D is correct. Parallel lines would have proportional x and y coefficients but different constants, intersecting lines would have different slopes, and perpendicularity is not indicated here. Scalar simplification proves coincidence directly.

Multiplying the first equation by (2) gives (4x+10y=30), different from the second. Same coefficient ratio with different constant gives parallel lines.

The direct answer is A: \((2,-1)\). A point of intersection must satisfy both equations. Test A: with \(x=2,y=-1\), the first left side is \(x-2y=2-2(-1)=2+2=4\), and the second is \(2x-y=2(2)-(-1)=4+1=5\). Both equations hold, so A is correct. B, \((-2,1)\), gives \(-2-2(1)=-4\), not 4. C, \((1,-2)\), gives \(1-2(-2)=5\), not 4, although its second value is 4-? actually \(2(1)-(-2)=4\), not 5. D, \((3,-1)\), gives \(3-2(-1)=5\), not 4, and \(6+1=7\), not 5. Thus only A satisfies both. The negative y-coordinate is important: subtracting a negative number becomes addition. Memory cue: substitute both coordinates into both equations and check both results.

To find the x-intercept, put y = 0: 3x = 12, so x = 4 and the point is (4, 0). To find the y-intercept, put x = 0: 2y = 12, so y = 6 and the point is (0, 6). Therefore, option B is correct. Exam tip: at the x-intercept y is zero, while at the y-intercept x is zero; option C uses the wrong values for both intercepts.

Substituting \(x=0\) gives \(4(0)-3y=12\), so \(-3y=12\) and \(y=-4\). Therefore, the point is \((0,-4)\). The option \((0,4)\) results from missing the negative sign. Exam tip: setting \(x=0\) gives the point where the line intersects the y-axis.

The governing concept is graphical solution of a pair of linear equations: the coordinates of the intersection point must satisfy both equations simultaneously. Substitute x=5 and y=1 into each option. For option B, x+y=5+1=6 and x-y=5-1=4, so both equations are true. Therefore (5,1) lies on both lines represented by option B, making it their common graphical solution. In option A, the two required values are interchanged; option C uses 5 and 1 as separate results rather than the sum and difference; and option D gives 7 and 3, which do not match the point. Hence B is the only unambiguous answer.

Checking ( (4,3) ) gives (2(4)+3(3)=17), not (18). Correct calculation gives (x=3) and (y=2).

Putting ( (3,2) ) gives (2(3)+3(2)=12), so it is not correct. The correct solution is ( (\frac{21}{5},\frac{16}{5}) ), so recalculation is needed in such options.

The governing concept is that the graphical solution of two linear equations is their common intersection point. The coordinates must satisfy both equations. Test option A: for (4,3), the first equation gives x+2y=4+2(3)=10, and the second gives 2x−y=2(4)−3=5. Thus (4,3) lies on both lines. The other options fail this simultaneous check: (3,4) gives 11 and 2; (5,2) gives 9 and 8; and (2,5) gives 12 and −1. Since only one point satisfies both equations exactly, the lines meet at (4,3). Algebraically, solving the pair also confirms this intersection, so option A is unambiguous.

( (5,2) ) satisfies both equations. In exams, quickly check options by substituting them in both equations.

The intersection point is the common solution of both linear equations. Substituting (2, 3) gives \(2(2)+3(3)=4+9=13\) and \(2+3=5\), so this point lies on both lines. The other options satisfy \(x+y=5\), but not \(2x+3y=13\). Exam tip: verify a graphical intersection point by substituting its coordinates into both equations.

Substituting ( (4,3) ) gives \(4(4)-3=13\), so checking is necessary. The correct solution is \( \left(\frac{18}{5},\frac{17}{5}\right) \).

Adding \(x+y=7\) and \(4x-y=11\) gives \(5x=18\). Hence \(x=\frac{18}{5}\) and \(y=\frac{17}{5}\).

The governing concept is finding an axis intercept from a linear equation. Every point on the y-axis has x-coordinate 0, so substitute x=0 into 3x−2y=12. This gives 3(0)−2y=12, hence −2y=12 and y=−6 after division by −2. Therefore the y-intercept is (0,−6), making option B correct. The sign matters: substituting (0,6) gives −12 on the left, not 12, so option A fails. If y=0 is substituted instead, then 3x=12 and x=4, giving (4,0), the x-intercept represented by option C. Option D has the wrong axis intercept and also does not satisfy the equation. Checking the selected point directly confirms 3(0)−2(−6)=12.

Multiplying the first equation by (2) gives (2x-2y=4), which is different from the second. Hence the lines are parallel and have no solution.

The direct answer is B: coincident lines. Divide every term of 3x+6y=18 by 3. This gives x+2y=6, exactly the second equation. Hence both graphs are the same straight line and every point on it is a common solution. A is wrong because parallel and distinct lines have equal direction but different positions and no common point; these equations have identical positions. B is correct because the equations are equivalent. C is wrong because one intersection occurs when two non-identical lines cross at one point. D is wrong because perpendicular lines have slopes whose product is −1 and meet at a right angle; these lines overlap instead. For a quick coefficient check, the first equation has all parts three times the second: 3/1=3, 6/2=3, and 18/6=3. Thus the ratios match completely, giving infinitely many solutions. Memory cue: divide the larger-looking equation by a common factor before deciding the graph type.

When \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\), the two lines have the same slope but different intercepts. Therefore, they are distinct parallel lines and have no common solution, so the pair is inconsistent. Exam tip: if all three ratios are equal, the lines are coincident.

For the pair a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the condition a₁/a₂ ≠ b₁/b₂ means that the corresponding coefficients are not proportional in the same way. Consequently, the two lines have different slopes. Two distinct straight lines with different slopes must intersect at one and only one point, and that common point is the unique ordered-pair solution of the pair. Therefore option B is correct. Such a pair is called consistent and independent. Option A applies to distinct parallel lines, where a₁/a₂ = b₁/b₂ but c₁/c₂ is different. Option C applies to coincident lines, where all three ratios are equal. Two ordinary straight lines cannot have exactly two distinct intersection points, so option D is impossible.

Multiplying the first equation by (2) gives (4x+6y=14), while the second is (4x+6y=20). Hence the lines are parallel and distinct.

Substituting (3, 0) into both equations gives 2(3) − 0 = 6 and 3 + 0 = 3. Therefore, the two lines intersect at (3, 0), which is the graphical solution. For example, (2, 1) satisfies the second equation, but in the first equation it gives 4 − 1 = 3, not 6. Exam tip: always verify the intersection point in both equations.

The intersection point must satisfy both equations. From the first equation, \(x=4\). Substituting this into the second equation gives \(2(4)+y=11\), so \(y=3\). Therefore, the intersection point is \((4,3)\). Option B reverses the coordinates and does not satisfy \(x=4\). Exam tip: Always substitute the proposed point into both equations to verify it.

The intersection point must satisfy both equations. From the first line, \(y=2\). Substituting this into the second equation gives \(3x+2=14\), so \(3x=12\) and \(x=4\). Therefore, the intersection point is \((4,2)\). In option A, the x- and y-coordinates are interchanged. Exam tip: verify an intersection point by substituting both coordinates into both equations.