Graphical method of finding solutions.

The second equation is (2) times the first, so the lines are coincident. Coincident lines have infinitely many solutions.

Adding the equations gives (2x=16), so (x=8) and (y=4). In a real problem, the meeting point is the graphical solution.

The governing concept is the intersection of two linear equations: the common point must satisfy both equations simultaneously. Subtract the second equation from the first: (2x + 3y) − (x + 3y) = 21 − 15, so x = 6. Substitute this value into x + 3y = 15: 6 + 3y = 15, hence 3y = 9 and y = 3. Therefore the lines meet at (6, 3), so option B is correct. Option A gives 2(5)+3(3)=19, not 21; option C reverses the coordinates; and option D gives 2(6)+3(2)=18, so those cannot be the intersection.

The intersection point is the simultaneous solution of both linear equations. Subtract the first equation from the second: (2x + 4y) − (x + 4y) = 20 − 18, which gives x = 2. Substituting x = 2 into x + 4y = 18 gives 2 + 4y = 18, so 4y = 16 and y = 4. Thus the lines meet at (2, 4), making option A correct. A quick check confirms it: 2 + 4(4) = 18 and 2(2) + 4(4) = 20. Option B swaps the coordinates, option C fails the first equation, and option D gives 22 in the first equation rather than 18.

In \( \left(-3,2\right) \), the first coordinate is (x) and the second is (y). Do not change order with negative coordinates.

The intersection point must satisfy both linear equations simultaneously. For option A, 2(2) - (-1) = 4 + 1 = 5 and 2 + 2(-1) = 0, so both equations hold. The remaining options fail at least one equation when substituted. Hence the two graphs meet at (2, -1), making option A the unique answer.

Substituting the point \\(1,5\\) in option A gives \\(2(1)+5=7\\), and substituting \\(3,1\\) gives \\(2(3)+1=7\\). Thus, both points satisfy \\(2x+y=7\\), so this equation represents their line. For example, option C works for the first point because \\(1+5=6\\), but it fails for the second point since \\(3+1=4\\neq6\\). Exam tip: The quickest method is to substitute both given points into each option.

On the line x = 0, the value of x is zero. Substituting x = 0 in the first equation gives 3y = 12, so y = 4. Therefore, the intersection point is (0, 4). The point (4, 0) is the x-intercept, obtained by setting y = 0. In exams, remember that x = 0 represents the y-axis, while y = 0 represents the x-axis.

The line y = 0 represents the x-axis. To find the intersection, substitute y = 0 in 4x + y = 20: 4x = 20, so x = 5. Therefore, the intersection point is (5, 0). In (0, 5), the coordinates are interchanged, while (4, 0) does not satisfy the equation. Exam tip: y = 0 represents the x-axis, whereas x = 0 represents the y-axis.

Completely overlapping lines are coincident lines. At every point on them, both equations are true.

Subtracting the equations gives (x=5), then (5+y=9) gives (y=4). In real life, the meeting point is the intersection.

The intersection point is the common solution of both equations. Adding the equations gives \(3x=12\), so \(x=4\). Substituting this into \(x-3y=-6\) gives \(4-3y=-6\), hence \(y=\frac{10}{3}\). Therefore, the intersection is \(\left(4,\frac{10}{3}\right)\). Option B incorrectly reverses the coordinates. Exam tip: After reading an intersection point from a graph, verify it in both equations.

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

A point where two graphed lines meet must satisfy both equations. Subtract the second equation from the first: (3x + 2y) − (x + 2y) = 19 − 9, giving 2x = 10 and therefore x = 5. Substitute x = 5 into x + 2y = 9: 5 + 2y = 9, so 2y = 4 and y = 2. Hence the graphical intersection is (5, 2), so option A is correct. Verification gives 3(5)+2(2)=19 and 5+2(2)=9. Option B reverses the coordinates, while options C and D fail at least one of the two original equations.

To find the \(y\)-intercept, put \(x=0\). Then \(7(0)+2y=28\), so \(y=14\). Therefore, the line cuts the \(y\)-axis at \((0,14)\). Option B is the \(x\)-intercept because it is obtained by putting \(y=0\). Exam tip: For a \(y\)-intercept, always set \(x=0\).

To find the x-intercept, put \(y=0\). Then \(2x-5(0)=20\), so \(2x=20\) and \(x=10\). Therefore, the x-intercept is \((10,0)\). Option D has an incorrect x-coordinate, while options A and C lie on the y-axis. Exam tip: Set \(y=0\) to find the x-intercept.

Dividing the first equation by (5) gives (x+2y=6). Hence both lines are coincident and give infinitely many solutions.

Multiplying the first equation by (2) gives (4x+6y=22), different from the second. Therefore the lines are parallel and have no solution.

The intersection of two lines is the ordered pair that satisfies both line equations. Since the terms involving y are identical, subtracting one equation from the other removes 5y and makes the calculation simple. Starting with 2x+5y=16 and subtracting x+5y=13 gives x=3.

Now substitute x=3 into the first equation: 3+5y=13. Therefore 5y=10 and y=2. The lines meet at (3,2), so option A is correct. Checking the second equation gives 2(3)+5(2)=6+10=16, confirming the result. A point such as (2,3) may look similar because its coordinates are reversed, but it gives 2+15=17 in the first equation and therefore is not the intersection.

At ( (4,2) ), (4(4)+2=18), but (4+2=6). For a common solution, the point must satisfy both equations.