Graphical method of finding solutions.

Correct answer: A. Point \(\left(3,4\right)\)

Substituting \(\left(3,4\right)\) gives \(2\left(3\right)+4=10\) and \(3+2\left(4\right)=11\). This is the common point of both lines.

Correct answer: A. Point \(\left(\frac{42}{11},\frac{67}{11}\right)\)

From the first equation, \(y=5x-13\). Substituting this in the second equation gives \(x+2(5x-13)=16\), so \(11x=42\) and \(x=\frac{42}{11}\). Therefore, \(y=5\left(\frac{42}{11}\right)-13=\frac{67}{11}\). Hence, the graphs intersect at \(\left(\frac{42}{11},\frac{67}{11}\right)\). Option B incorrectly interchanges the \(x\)- and \(y\)-coordinates. Exam tip: Always substitute the intersection coordinates into both original equations to verify them.

Correct answer: A. \\(2x+y=5\\)

Substitute both points into the options. For \\((-1,7)\\), \\(2(-1)+7=5\\), and for \\( (2,1)\\), \\(2(2)+1=5\\). Thus both points satisfy \\(2x+y=5\\). Alternatively, the slope is \\(m=(1-7)/(2-(-1))=-2\\), giving \\(y=-2x+5\\), or \\(2x+y=5\\). Exam tip: To verify a line equation, substitute both given points and check that each one satisfies it.

Correct answer: C. They are parallel and distinct

Dividing the second equation by (2) gives (x+2y=10). Same left side with different constants gives parallel lines.

Correct answer: C. The intersection point is \(\left(\frac{7}{3},\frac{5}{3}\right)\)

Fractional coordinates need more care when read from a graph. In such questions, the scale must be very clear.

Correct answer: B. Point \(\left(4,\frac{11}{3}\right)\)

Subtracting the second from the first gives \(4y=12\), so \(y=3\); careful checking is needed. The correct solution is \(\left(5,3\right)\).

Correct answer: A. Point \(\left(5,3\right)\)

Substituting \(\left(5,3\right)\) gives \(2\left(5\right)+3\left(3\right)=19\) and \(2\left(5\right)-3=7\). If both are true, this is the solution.

Correct answer: A. 1

The point \((2,6)\) satisfies the first equation because \(2+6=8\). Substituting \(x=2\) and \(y=6\) into the second equation gives \(2k+2(6)=14\), or \(2k+12=14\). Thus, \(2k=2\) and \(k=1\). Therefore, option A is correct. Exam tip: To test whether a point lies on a line, substitute its coordinates directly into the equation.

Correct answer: B. \(\frac{12}{5}\)

The intersection point \((2,5)\) is a common solution of both lines. Substituting \(x=2\) and \(y=5\) in the first equation gives \(2(2)+a(5)=16\), or \(4+5a=16\). Hence, \(5a=12\) and \(a=\frac{12}{5}\). Option A is incorrect because it makes the left-hand side equal to \(14\), not \(16\). Exam tip: When a graphical solution is given, substitute the point into both equations to verify it.

Correct answer: A. \(2x+4y=18\)

In option A, dividing every term of \(2x+4y=18\) by 2 gives \(x+2y=9\). Hence both equations have exactly the same graph and represent coincident lines. Options B and C produce distinct lines with the same slope, while option D has a different slope. Exam tip: for coincident lines, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Correct answer: C. (4x-6y=18)

Dividing (4x-6y=18) by (2) gives (2x-3y=9). Same left side with different constants gives distinct parallel lines.

Correct answer: A. Point \(\left(4,3\right)\)

Subtracting the equations gives \(2x=8\), so \(x=4\) and \(y=3\). In a real situation this is the meeting point.

Correct answer: A. Point \(\left(8,\frac{13}{5}\right)\)

Subtracting the equations gives \(x=8\), then \(8+5y=21\) gives \(y=\frac{13}{5}\). This is the graphical intersection.

Correct answer: A. Point \(\left(\frac{21}{5},\frac{23}{5}\right)\)

Solving both equations gives \(x=\frac{21}{5}\) and \(y=\frac{23}{5}\). This point lies on both lines.

Correct answer: A. 7

The intersection point \((3,1)\) lies on both lines. Substituting \(x=3\) and \(y=1\) into the first line gives \(3+a(1)=10\), so \(a=7\). The second equation also checks correctly: \(2(3)-1=5\). Exam tip: Substitute the given point into the equation containing the unknown parameter.

Correct answer: B. 3

The intersection point must satisfy both line equations. The second equation is verified because \(2+4=6\). Substituting \(x=2\) and \(y=4\) into the first equation gives \(2k+2(4)=14\), so \(2k+8=14\), \(2k=6\), and hence \(k=3\). Exam tip: Substitute the coordinates of the given intersection point into the equation containing the unknown parameter.

Correct answer: A. Error of sign and coordinate order

In \(\left(6,-2\right)\), \(x=6\) and \(y=-2\). Reversing coordinates and changing sign makes the answer wrong.

Correct answer: A. Point \(\left(4,8\right)\)

Subtracting the equations gives \(2x=8\), so \(x=4\) and \(y=8\). Whatever the context, the intersection point is the solution.

Correct answer: B. 3

For two lines to be coincident, the ratios of their corresponding coefficients must be equal. Writing both equations in standard form gives \(\frac{2}{2}=\frac{a}{3}=\frac{18}{18}\). Hence \(\frac{a}{3}=1\), so \(a=3\). Option 2 refers only to the coefficient of \(x\), so it is not correct. Exam tip: for coincident lines, use \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Correct answer: A. (3)

The direct answer is A: \(b=3\). First simplify the second line: divide \(2x+6y=18\) by 2 to obtain \(x+3y=9\). The first line is \(x+by=7\). For the two lines to be parallel, the coefficients of x and y must have the same ratio, so b must be 3. With b=3, the equations become \(x+3y=7\) and \(x+3y=9\). They have identical left sides but different constants, so they cannot be the same line and never meet; hence they are parallel and distinct. Option A is correct. Option B, b=2, gives a different coefficient ratio and generally produces intersecting lines. Option C, b=6, also gives a different slope. Option D, b=1, likewise does not match the required direction. Notice that b=3 does not make the lines coincident because 7 is not 9. Exam cue: same x-y coefficient ratio but unequal constants means parallel distinct lines.

Correct answer: A. Point \(\left(0,-3\right)\)

Substituting \(\left(0,-3\right)\) gives \(3\left(0\right)-4\left(-3\right)=12\). A negative (y)-intercept is plotted downward on the graph.

Correct answer: A. (6)

Here \(x+y=\frac{5}{2}+\frac{7}{2}=\frac{12}{2}=6\). Values of (x) and (y) are read directly from the intersection point.

Correct answer: A. (1)

For parallel lines, (\frac{k}{2}=\frac{3}{6}), so (k=1). The constants ratio is different, so the lines are distinct parallel lines.

Correct answer: A. Point \(\left(\frac{21}{5},\frac{16}{5}\right)\)

Using \(x=y+1\) from \(x-y=1\) gives \(y=\frac{16}{5}\) and \(x=\frac{21}{5}\). On the graph this is the intersection point.

Correct answer: A. Point: \(\left(\frac{25}{7},\frac{37}{7}\right)\)

From the first equation, \(y=4x-9\). Substituting this in the second equation gives \(2x+3(4x-9)=23\), so \(14x=50\) and \(x=\frac{25}{7}\). Therefore, \(y=4\left(\frac{25}{7}\right)-9=\frac{37}{7}\). Hence, the intersection point of the two lines is \(\left(\frac{25}{7},\frac{37}{7}\right)\). In option B, the coordinates are interchanged, so the point does not satisfy the equations. Exam tip: Always substitute the intersection point into both original equations to verify it.