Graphical method of finding solutions.

Correct answer: A. Point \(\left(5,2\right)\)

Substituting \(\left(5,2\right)\) gives \(5\left(5\right)+3\left(2\right)=31\) and \(5+2=7\). If both equations are true, that is the intersection.

Correct answer: A. Point \(\left(\frac{91}{17},\frac{50}{17}\right)\)

Elimination gives \(17y=50\) and \(x=\frac{91}{17}\). Fraction coordinates can also be graphical solutions.

Correct answer: B. Parallel and distinct lines

(\frac{a_1}{a_2}=\frac{b_1}{b_2}=3), but (\frac{c_1}{c_2}=\frac{12}{5}). Hence the lines are parallel and inconsistent.

Correct answer: C. Coincident

The direct answer is option C: the graphs are coincident, meaning they are the same line. Compare the equations 8x-12y=20 and 2x-3y=5. Dividing every term of the first equation by 4 gives 2x-3y=5, exactly the second equation. Multiplying or dividing a whole linear equation by a nonzero number does not change its graph. Therefore every point on one line lies on the other, so there are infinitely many common solutions. A, parallel and distinct, is wrong because distinct parallel lines have different equations after simplification and no common point. B, intersecting at one point, is wrong because that would produce one common solution, not identical equations. D, intersecting at the origin, is wrong because the origin (0,0) does not satisfy 2x-3y=5; its left side is 0, not 5. Coincident lines are visually one line drawn twice. Memory cue: simplify both equations first; identical simplified equations mean coincident lines.

Correct answer: A. \\( (5,0) \\) and \\( (0,-9) \\)

To find the x-intercept, put y=0: 9x=45, giving x=5 and the point \\( (5,0) \\). To find the y-intercept, put x=0: -5y=45, giving y=-9 and the point \\( (0,-9) \\). Therefore, option A is correct. Option C incorrectly treats 9 and 5 as the coordinates instead of calculating the intercepts. Exam tip: set y=0 for the x-intercept and x=0 for the y-intercept.

Correct answer: A. Point \(\left(-5,-9\right)\)

Putting \(x=-5\) gives \(4\left(-5\right)-3y=7\), so \(y=-9\). In a vertical line, (x) is already fixed.

Correct answer: A. Point \((7,6)\)

At the intersection, both equations must hold. Since \(y=6\), substitute 6 for \(y\) in the second equation: \(5x-2(6)=23\), so \(5x=35\) and \(x=7\). Thus, the intersection point is \((7,6)\).

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

The intersection point on the graph must satisfy both equations simultaneously. Adding the two equations gives \(5x=25\), so \(x=5\). Substituting this into the first equation gives \(5+3y=14\), hence \(y=3\). Therefore, the intersection point is \((5,3)\). Exam tip: verify a suspected ordered pair in both equations before selecting it.

Correct answer: A. Point \(\left(\frac{25}{7},\frac{23}{7}\right)\)

\(\left(\frac{25}{7},\frac{23}{7}\right)\) satisfies both equations. Read fraction coordinates carefully using the graph scale.

Correct answer: A. Point \(\left(2,5\right)\)

At \(\left(2,5\right)\), \(3\left(2\right)+4\left(5\right)=26\), but \(2+5=7\) also, so check fully. The correct non-common point is \(\left(4,\frac{7}{2}\right)\).

Correct answer: B. \(x=-\frac{5}{2},\ y=3\)

The first coordinate of a point is (x) and the second is (y). Do not change order while reading negative fraction coordinates.

Correct answer: B. They are parallel and distinct

Multiplying the first equation by (2) gives (10x-6y=20), while the second is (10x-6y=25). Hence the lines are parallel and distinct.

Correct answer: C. Infinitely many solutions

Every point on coincident lines satisfies both equations, so the system has infinitely many solutions. Check \(a_1/a_2=b_1/b_2=c_1/c_2\). Distinct parallel lines have no solution. Exam tip: identify the line relationship before selecting the answer.

Correct answer: B. Point \((0,-6)\)

The line \(x=0\) represents the y-axis, so substitute \(x=0\) into the first equation: \(6(0)-7y=42\), giving \(-7y=42\) and hence \(y=-6\). Therefore, the intersection point is \((0,-6)\). Option A has the wrong sign for the y-coordinate, while option C is the x-intercept of the first line and does not lie on \(x=0\). Exam tip: when \(x=0\), the first coordinate of the point must be zero.

Correct answer: C. Lines are parallel

(\frac{a_1}{a_2}=\frac{b_1}{b_2}=3), but (\frac{c_1}{c_2}=\frac{11}{4}). Hence the lines are parallel and inconsistent.

Correct answer: A. \(\left(0,-4\right)\) and \(\left(16,0\right)\)

At \(x=0\), \(y=-4\), and at \(y=0\), \(x=16\). While finding intercepts, note which variable is kept zero.

Correct answer: A. \(\left(\frac{15}{4},-\frac{5}{2}\right)\)

\(3.75=\frac{15}{4}\) and \(-2.5=-\frac{5}{2}\). It is better to convert decimal coordinates into simplified fractions.

Correct answer: A. Point \(\left(4,5\right)\)

Substituting \(\left(4,5\right)\) gives \(3\left(4\right)+5=17\) and \(4+3\left(5\right)=19\). This is the common point of both lines.

Correct answer: A. Point \(\left(\frac{79}{19},\frac{113}{19}\right)\)

From the first line, \(y=6x-19\). Substituting this into the second equation gives \(x+3(6x-19)=22\), so \(19x=79\) and \(x=\frac{79}{19}\). Therefore, \(y=6\left(\frac{79}{19}\right)-19=\frac{113}{19}\). Hence, the intersection point is \(\left(\frac{79}{19},\frac{113}{19}\right)\). In option B, the values of \(x\) and \(y\) are interchanged. Exam tip: Substitute the ordered pair into both original equations to verify the solution.

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

Substituting the first point in \(2x+y=5\) gives \(2(-2)+9=5\), and substituting the second point gives \(2(3)+(-1)=5\). Thus, both points lie on this line. Options B and D satisfy only one of the points, while option C does not correctly satisfy either given point. Exam tip: To verify the equation of a line, substitute both given ordered pairs into it.

Correct answer: C. They are parallel and distinct

Dividing the second equation by (2) gives (2x+3y=15). Same left side with different constants gives parallel lines.

Correct answer: A. \((5,4)\)

The point of intersection is the common solution of both linear equations. Subtracting the second equation from the first gives \((3x+4y)-(3x-y)=31-11\), so \(5y=20\) and hence \(y=4\). Substituting \(y=4\) into \(3x-y=11\) gives \(3x-4=11\), which leads to \(x=5\). Therefore, the intersection point is \((5,4)\). As an exam tip, verify the ordered pair in both equations before selecting the answer.

Correct answer: A. \(\frac{10}{3}\)

The intersection point \(\left(4,3\right)\) satisfies both equations. Substituting \(x=4\) and \(y=3\) into the first equation gives \(3(4)+a(3)=22\), or \(12+3a=22\). Thus, \(3a=10\), so \(a=\frac{10}{3}\). Option D results from dividing \(22\) by \(3\) without first subtracting the \(12\) contributed by \(3x\). Exam tip: Substitute the coordinates of the graphical intersection point into the equation containing the unknown parameter.

Correct answer: C. (6x-4y=20)

Dividing (6x-4y=20) by (2) gives (3x-2y=10). Same left side with different constants gives distinct parallel lines.

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

Subtracting the equations gives \(3x=15\), so \(x=5\) and \(y=3\). In a real situation this is the meeting point.