Graphical method of finding solutions.

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

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.

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.

The governing concept is that every ordered pair in a value table for a line must satisfy the given linear equation. For option A, substitute the first pair: 5(2) + 11 = 10 + 11 = 21. For the second pair, 5(4) + 1 = 20 + 1 = 21. Thus both points lie on the line, so A is correct. In B, the substitutions give 11 and 31, not 21. In C, 5(0) + 5 = 5 and 5(5) + 21 = 46. In D, the results are 26 and 24. Therefore only A gives two valid points for graph construction.

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.

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

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.

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.

The governing concept is equivalence of linear equations. Two equations represent coincident lines when all three corresponding coefficients, including the constant term, are multiplied by the same non-zero number. Multiplying 2x + 5y = 13 by 2 gives 4x + 10y = 26, exactly the equation in option A. Dividing that equation by 2 returns the original equation, so both equations have every point in common and produce the same graph. Option B has the wrong constant, so it represents a parallel line. Option C keeps the left side but changes the constant, also producing a distinct parallel line. Option D changes the coefficient of y and is not an equivalent equation. Hence A is the only correct choice.

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

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

Subtracting the equations gives \(2x=14\), then \(x=7\) and \(7+5y=25\) gives \(y=\frac{18}{5}\). This is the graphical intersection.

Solving both equations gives \(y=\frac{22}{5}\) and \(x=\frac{26}{5}\). \(\frac{52}{10}\) can also be written as \(\frac{26}{5}\).

The intersection point \((2,4)\) must satisfy both line equations. Substituting \(x=2\) and \(y=4\) into the first equation gives \(2k+4(4)=22\), or \(2k+16=22\). Hence, \(2k=6\) and \(k=3\), so option B is correct. As a check, the point also satisfies the second equation because \(2+4=6\). Exam tip: when the intersection point is given, substitute its coordinates into the equation containing the unknown parameter.

In \(\left(7,-3\right)\), \(x=7\) and \(y=-3\). Reversing coordinates and changing sign makes the answer wrong.

Subtracting the equations gives \(3x=12\), so \(x=4\) and \(y=10\). Whatever the context, the intersection point is the solution.

Dividing (3x+12y=30) by (3) gives (x+4y=10). Thus (b=4) makes the lines parallel and distinct.

Substituting \(\left(0,-5\right)\) gives \(5\left(0\right)-6\left(-5\right)=30\). A negative (y)-intercept is plotted downward on the graph.

Here \(x+y=\frac{7}{2}+\frac{9}{2}=\frac{16}{2}=8\). Values of (x) and (y) are read directly from the intersection point.

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