Graphical method of finding solutions.

The point of intersection must satisfy both equations simultaneously. For (2, 0), x + y = 2 + 0 = 2 and x − y = 2 − 0 = 2, so it is the correct point. The point (0, 2) satisfies the first equation but not the second. Exam tip: substitute the coordinates of each option into both equations.

Both equations are exactly the same, so they represent the same line. Such a pair has infinitely many solutions.

Dividing the whole equation by (2) gives (2x+y=4). Proportional equations can often give coincident lines.

The solution is obtained from the coordinates of the intersection point. Intercepts alone are not enough when a second line is given.

Substituting x = 5 in the equation gives 5 + y = 8, so y = 3. Therefore, the required point is (5, 3). In option B, the x- and y-values are written in the reverse order, whereas a point is always written as (x, y). Exam tip: Verify a point by substituting both coordinates into the equation.

Substituting \(y=1\) gives \(x-1=3\), so \(x=4\). Therefore, the point is \((x,y)=(4,1)\). Option B reverses the coordinate order; a point is written as \((x,y)\), not \((y,x)\). Exam tip: find the missing coordinate first, then write the ordered pair as \((x,y)\).

Substituting \(x=0\) gives \(2(0)+3y=12\), so \(3y=12\) and \(y=4\). Therefore, the point is \((0,4)\), which is also the \(y\)-intercept of the line. In option B, the values of \(x\) and \(y\) are interchanged, while C and D do not satisfy the equation. Exam tip: setting \(x=0\) gives the point where the line meets the \(y\)-axis.

Putting \\(y=0\\) in the equation gives \\(3x+2(0)=12\\), so \\(3x=12\\) and \\(x=4\\). Therefore, the point is \\((4,0)\\). Option B is obtained by putting \\(x=0\\), so it represents the y-intercept instead. Exam tip: To find the x-intercept, set \\(y=0\\).

The solution must be read in ( (x,y) ) order, and reversing it changes the answer. In exams, write the intersection point carefully.

The solution is the coordinates of the common point where both lines meet. In exams, always write the point in ( (x,y) ) order.

The governing concept is obtaining a coordinate point on a linear equation by assigning one coordinate and calculating the other. Substitute x = 4 into x + y = 11: 4 + y = 11. Subtracting 4 from both sides gives y = 7. Therefore the ordered pair is (x, y) = (4, 7), so option A is correct. Option B reverses the x- and y-coordinates and would give 7 + 4 = 11 only because the equation is symmetric, but it does not satisfy the stated condition x = 4. Options C and D use 11 as one coordinate without solving correctly. The point can be plotted as an ordinary point on the line.

Substituting \(x=1\) gives \(2(1)+y=10\), or \(2+y=10\). Therefore, \(y=8\). Choosing 6 would make the left side equal to 8, not 10, so it is incorrect. Exam tip: to find a point on a linear equation, substitute the known coordinate and solve for the other variable.

To find the y-intercept, put x=0 in the equation. Then \(3(0)+y=15\), so y=15 and the point is \((0,15)\). Option A is the x-intercept because it is obtained by putting y=0. Exam tip: every point on the y-axis has x-coordinate 0.

The \(x\)-intercept is the point where the line crosses the \(x\)-axis, so \(y=0\). Substituting this into the equation gives \(4x=20\), hence \(x=5\). Therefore, the \(x\)-intercept is \((5,0)\). Remember that \((0,20)\) is the \(y\)-intercept because \(x=0\) there.

Both have the same coefficients of (x) and (y), but different constants. Hence, the lines are parallel and have no solution.

The second equation is (3) times the first, so both are the same line. In this case, there are infinitely many solutions.

In the graphical method, the point of intersection of two lines is the common solution of both equations. Since the intersection point is \((6,2)\), we have \(x=6\) and \(y=2\), so option D is correct. Option A incorrectly reverses the coordinates. Exam tip: In \((x,y)\), the first coordinate is always \(x\) and the second is \(y\).

Substituting ( (2,5) ) gives (2+5=7), so (x+y=7) is correct. To check a point, substitute it in the equation.

Putting (x=3) gives (y=3+4=7), so ( (3,7) ) is correct. To check a point on a line, substitute the value of (x).

Substituting \(x=2\) gives \(y=3(2)-1=6-1=5\). Therefore, the point is \((x,y)=(2,5)\). In option A, the value of \(y\) is incorrectly taken as 3. In the exam, remember to write coordinates in the order \((x,y)\).