Graphical method of finding solutions.

A single common point means that the pair has exactly one unique solution. Therefore, it is consistent and independent. In an inconsistent pair, the lines are parallel and have no solution, whereas a consistent dependent pair has coincident lines and infinitely many solutions. Exam tip: one point of intersection means one unique solution, so the pair is consistent and independent.

Coincident lines lie exactly on top of each other, so every point on the common line is a solution of both equations. Hence, the pair has infinitely many solutions and is called consistent and dependent. Option C is incorrect because an independent pair has lines that intersect at exactly one point. Exam tip: coincident lines imply infinitely many solutions and a consistent dependent pair.

For the ordered pair \((7, 3)\), \(x+y=7+3=10\) and \(x-y=7-3=4\). Thus, it satisfies both equations, so \((7,3)\) is the intersection point of the two lines on the graph. For example, \((6,4)\) satisfies the first equation but \(6-4=2\), not 4. Exam tip: substitute the coordinates into both equations to verify a solution.

Subtracting the second equation from the first gives \\(x=2\\). Substituting this into \\(x+y=5\\) gives \\(y=3\\). Therefore, the intersection point is \\(2,3\\). For example, option B gives \\(2(3)+2=8\\), so it does not satisfy the first equation. Exam tip: verify a graphical-solution point in both equations.

A solution of two linear equations is an ordered pair whose first number is the value of x and whose second number is the value of y. The same pair must make both equations true at the same time. This is also the point where the two corresponding lines intersect on a graph.

For option A, take x=3 and y=2. In the first equation, x+3y becomes 3+3(2)=3+6=9, so it is satisfied. In the second equation, x+y becomes 3+2=5, so that equation is also satisfied. Therefore, the common solution is (3,2). A pair that satisfies only one equation would not be a solution of the pair.

The direct answer is option A: the origin, (0,0). The x-axis is the horizontal line whose points have y = 0, and the y-axis is the vertical line whose points have x = 0. A point common to both axes must therefore have x = 0 and y = 0. Hence their intersection is (0,0), called the origin. Option A is correct. Option B, (1,1), lies away from both axes because neither coordinate is zero. Option C, (0,1), lies on the y-axis, but not on the x-axis because y is not zero. Option D, (1,0), lies on the x-axis, but not on the y-axis because x is not zero. Thus only (0,0) belongs to both axes. Memory cue: both axes meet at zero-zero.

To find the x-intercept, put \(y=0\), giving \(x=6\) and the point \((6,0)\). To find the y-intercept, put \(x=0\), giving \(y=6\) and the point \((0,6)\). Hence, A is correct. In option B, both coordinates are 3; although \(3+3=6\), those points are not on the respective axes. Exam tip: set \(y=0\) for the x-intercept and \(x=0\) for the y-intercept.

Putting \(x=0\) in \(2x+y=4\) gives \(y=4\), so \((0,4)\) lies on the line. Putting \(y=0\) gives \(2x=4\), hence \(x=2\), so \((2,0)\) is the second point. The points in option B do not satisfy the equation; for example, substituting \((4,0)\) gives a left side of 8. Exam tip: To obtain two convenient points, set \(x=0\) and \(y=0\) successively.

Substituting the coordinates of \((4,2)\) gives \(x-2y=4-2(2)=4-4=0\). Therefore, the point lies on the line. The other points do not make the left-hand side equal to zero. In such questions, verify a point by substituting its coordinates directly into the equation.

In an ordered pair \((x, y)\), the first value is the \(x\)-coordinate and the second value is the \(y\)-coordinate. Therefore, for \((1, 2)\), \(x=1\). Option 2 is the \(y\)-coordinate, not the \(x\)-coordinate. Exam tip: Always read coordinates in the order \((x, y)\).

In an ordered pair (x, y), the first coordinate is x and the second coordinate is y. Therefore, the y-coordinate of (5, 2) is 2. Option 5 is the x-coordinate, not the y-coordinate. Exam tip: Always read the coordinates in the order (x, y).

The equation \(x=0\) means that the \(x\)-coordinate of every point is zero, while the \(y\)-coordinate may have any value. All such points lie on the \(y\)-axis, so the correct answer is the \(y\)-axis. The \(x\)-axis is represented by \(y=0\), which is a useful exam tip to remember.

The equation \(y=0\) means that the y-coordinate of every point is zero. All such points lie on the x-axis, so this is the equation of the x-axis. The y-axis is represented by \(x=0\). Exam tip: a horizontal line has a constant y-value, while a vertical line has a constant x-value.

Both lines are parallel to the (y)-axis and have different (x)-values. Therefore, they are parallel and will not meet.

The equations \(y=1\) and \(y=4\) represent horizontal lines. Since their \(y\)-intercepts are different, they never meet and are parallel. They are not coincident because coincident lines have all points in common. Exam tip: the graph of \(y=c\) is always a horizontal line.

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.

The direct answer is option A, coincident lines. Both equations are exactly x+2y=4, so they have the same coefficients and the same constant. They do not describe two different lines; they describe one single line drawn twice. Every point satisfying the first equation also satisfies the second, so the pair has infinitely many common solutions. Option B, parallel lines, is wrong because distinct parallel lines have no common point, whereas these equations are identical. Option C, intersecting at one point, is wrong because two identical lines have infinitely many common points, not one. Option D is not a standard type of pair of lines and is irrelevant. Memory cue: identical equations mean coincident lines and 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.