Graphical method of finding solutions.

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.

To find the \(x\)-intercept, put \(y=0\): \(3x=24\), so \(x=8\) and the point is \((8,0)\). To find the \(y\)-intercept, put \(x=0\): \(4y=24\), so \(y=6\) and the point is \((0,6)\). Therefore, option A is correct; option B gives incorrect values, while C and D misidentify or reverse the intercept points. Exam tip: set \(y=0\) for the x-intercept and \(x=0\) for the y-intercept.

The y-intercept is the point where a line crosses the y-axis. Every point on the y-axis has \\(x=0\\). Substitute this value into \\(5x-3y=15\\): \\(5(0)-3y=15\\), so \\(-3y=15\\). Dividing by \\(-3\\) gives \\(y=-5\\). Therefore the y-intercept is \\((0,-5)\\), which is option B. The point \\((3,0)\\) is the x-intercept instead, since it is found by setting y equal to zero.

The result can also be checked by putting \\((0,-5)\\) into the equation: \\(5(0)-3(-5)=15\\), so the equation is satisfied. Because the y-coordinate is negative, the point lies five units below the origin on the y-axis. A positive point \\((0,5)\\) would give \\(-15\\), not 15, so it cannot be correct. Hence the supplied answer B is fully supported.

Dividing the first equation by (2) gives (x+2y=8). Therefore the lines are coincident and have infinitely many solutions.

Multiplying the first equation by (2) gives (2x-6y=10), different from the second. Hence the lines are parallel and have no solution.

This condition represents distinct parallel lines. Therefore there is no common point and the pair is inconsistent.

When all three ratios are equal, both equations represent the same line. In this case, infinitely many solutions are obtained.

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).

Since the first line is \(x=6\), the x-coordinate of the intersection must be 6. Substituting this in the second equation gives \(6+y=10\), so \(y=4\). Therefore, the intersection point is \((6,4)\). In \((6,10)\), the value of y has been incorrectly taken as 10. Exam tip: Always substitute the proposed point into both equations to verify it.

On the first line, \(y=-1\). Substituting this in the second equation gives \(2x-(-1)=9\), or \(2x+1=9\), so \(x=4\). Therefore, the point of intersection is \((4,-1)\). In option B the coordinates are reversed, while options C and D have an incorrect \(x\)-value or \(y\)-value, respectively. Exam tip: For a horizontal line of the form \(y=\text{constant}\), the y-coordinate of the intersection is known immediately.

Substituting the origin ( (0,0) ) gives left side (0), not (18). Check passing through origin by direct substitution.

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.

At ( (4,2) ), (3(4)+2(2)=16), but (4+2=6). To be a solution of both lines, both equations must be true.

\(3.5=\frac{7}{2}\) and \(2.5=\frac{5}{2}\). Write decimal coordinates read from a graph as simplified fractions.

Distinct parallel lines never meet, so there is no common point and hence no solution. Check \(a_1/a_2=b_1/b_2\ne c_1/c_2\). Exam tip: equal slopes with different intercepts indicate parallel lines.

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.

At (x=2), (6-y=4) gives (y=2), and at (x=3), (9-y=4) gives (y=5). Every table point must satisfy the equation.

Dividing the second equation by (2) gives (x+2y=8). Same left side with different constants gives parallel lines.

Here (\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac{1}{2}). Therefore both lines are coincident.

(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{1}{2}), but (\frac{c_1}{c_2}=\frac{9}{25}). Hence the lines are parallel and inconsistent.