Graphical method of finding solutions.

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 idea is that the x-intercept is found by putting y = 0, while the y-intercept is found by putting x = 0. For the x-intercept, 2x + 5(0) = 20, so 2x = 20 and x = 10; hence the point is (10, 0). For the y-intercept, 2(0) + 5y = 20, so 5y = 20 and y = 4; hence the point is (0, 4). Therefore option A is correct. Option B reverses the values, while options C and D do not satisfy the equation correctly or present the required pair accurately.

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.

Dividing the first equation by (3) gives (x+2y=6). Therefore both lines are the same and have infinitely many solutions.

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 two lines have different slopes. Lines with different slopes cannot remain parallel or coincide; they intersect at exactly one point. That common point gives the single ordered-pair solution of the equations. Thus option B is correct, and the pair is called consistent and independent. Option A describes parallel distinct lines, generally associated with equal coefficient ratios but an unequal constant ratio. Option C describes coincident lines, where all three ratios are equal, and option D is impossible for two distinct straight lines.

The governing concept is the relationship between the two equations. Multiplying x + 3y = 9 by 2 gives 2x + 6y = 18, which is exactly the second equation. Consequently, both equations represent the same straight line rather than two different lines. Every point lying on that line satisfies both equations, so there are infinitely many common solutions. Therefore option C is correct. Option A would apply to two lines that intersect at one point, option B to distinct parallel lines, and option D is not a possible number of intersections for two coincident or distinct straight lines in this context.

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.

Substituting the origin ( (0,0) ) gives left side (0), not (6). Check whether a line passes through origin by substitution.

For the point \((2,5)\), substituting \(x=2\) and \(y=5\) gives \(5x-2y=5(2)-2(5)=10-10=0\). Therefore, the point lies on the line. For example, substituting \((2,3)\) gives \(10-6=4\), so it does not lie on the line. In an exam, quickly test a point by substituting its coordinates into the equation.

( (2,2) ) gives (6) in both equations. The point lying on both lines is the intersection point.

\(2.5=\frac{5}{2}\) and \(1.5=\frac{3}{2}\). When reading decimals from a graph, write them as simplified fractions.

A point lies on a line when its coordinates satisfy the equation of that line. Therefore, for two lines to meet at the origin, the coordinates \((0,0)\) must satisfy both equations in the pair. This is a direct substitution test and does not require solving the whole system.

For option B, \(2x+y=0\) gives \(2(0)+0=0\), and \(x-y=0\) gives \(0-0=0\). Thus both lines pass through the origin, so their intersection is at the origin because they are distinct lines. In the other options, at least one equation has a nonzero constant that fails when \(x=y=0\). Hence option B is correct.

The second equation is exactly three times the first: multiplying \(x+2y=7\) by 3 gives \(3x+6y=21\). Hence both equations represent the same graph, so the lines are coincident. For parallel and distinct lines, the ratios of the coefficients of \(x\) and \(y\) are equal but the ratio of the constants is different. Exam tip: if \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\), the two lines are coincident.