Algebraic methods: Substitution method and Elimination method.

Substituting \(x=3\) gives \(2(3)-y=4\), or \(6-y=4\). Thus, \(-y=-2\), so \(y=2\). In the exam, handle the negative sign carefully; \(6\) is the value of \(2x\), not of \(y\).

The intersection point must satisfy both equations. Subtracting the second equation from the first gives \(4y=24\), so \(y=6\). Substituting this into \(4x-y=10\) gives \(4x-6=10\), hence \(x=4\). Therefore, the intersection point is \((4,6)\). Option B merely reverses the coordinates and does not satisfy both equations. Exam tip: always substitute the obtained point into both original equations to verify it.

The coordinates of a point lying on a line must satisfy the equation of that line. Substituting \(\left(4,5\right)\) into the second line, \(kx+3y=23\), gives \(4k+3(5)=23\), or \(4k+15=23\). Thus, \(4k=8\) and \(k=2\). The first equation is also verified because \(4+5=9\). Exam tip: To find an unknown parameter when a point lies on a line, substitute the point’s coordinates directly into the line equation.

In elimination, adding equations cancels coefficients with equal magnitude and opposite signs, such as \(3x\) and \(-3x\), since their sum is \(0\). Equal constants alone do not eliminate a variable. Exam tip: first align one variable’s coefficients.

The point of intersection must satisfy both equations simultaneously. Multiplying the equations by 2 and 5 respectively gives \(8x+10y=62\) and \(15x-10y=5\). Adding them yields \(23x=67\), so \(x=\frac{67}{23}\). Therefore, the x-coordinate of the intersection point on the graph is \(\frac{67}{23}\). The value \(3\) is incorrect because it does not satisfy the second equation. Exam tip: Always substitute the obtained coordinate into an original equation to verify a graphical answer.

From
(r-s=1)
, we get
(s=r-1)
. Substituting this into
(3r+s=19)
gives
(3r+r-1=19)
, so
(4r=20)
and
(r=5)
. Hence
(s=4)
and
(r+s=5+4=9)
. Therefore, option B is correct. Exam tip: the coordinates of the intersection point satisfy both line equations, so expressing one variable in terms of the other and substituting is the quickest method here.

The intersection point
(r,s)
lies on both lines, so its coordinates satisfy both equations. From
(r-s=1)
, we get
(s=r-1)
. Substituting this into the first equation gives
(4r+r-1=29)
, so
(5r=30)
and
(r=6)
. Hence
(s=5)
and
(r+s=6+5=11)
. Therefore, option C is correct. Exam tip: To find the intersection point, solve the two linear equations simultaneously.

Adding both equations gives (2x=6), so (x=3) and (y=2). In exams, add first when one variable cancels easily.

Given \(x=2\), substitute it in \(2x+y=7\): \(2(2)+y=7\), or \(4+y=7\). Hence, \(y=3\). If \(y=5\), the left side becomes \(2(2)+5=9\), not 7. Exam tip: after substitution, solve the resulting simple linear equation carefully.

Substituting (y=x+1) in (x+y=7) gives (2x+1=7), so (x=3) and (y=4). Use the already isolated variable in exams.

Subtracting the second equation from the first gives (2x=6), so (x=3). Subtract equal like terms when their signs are the same.

Adding both equations gives (2x=14), so (x=7) and (y=3). After finding one variable, substitute it in any equation.

Adding both equations gives (3x=9), so (x=3) and (y=5). Variables with opposite signs cancel quickly in elimination.

Using (x=2y) gives (2y+y=9), so (y=3) and (x=6). In substitution, form an equation in one variable.

Substituting (y=3x) gives (4x=8), so (x=2) and (y=6). Substitution is easiest when one variable is already expressed as a multiple.

Subtracting the second equation from the first gives (3x=9), so (x=3). Identify the common variable before subtracting.

From the second equation (x=y+2); substituting gives (3y+2=11), so (y=3) and (x=5). Substitute the isolated variable carefully.

Using (x=5-y) gives (10-2y+3y=12), so (y=2) and (x=3). It is better to isolate a variable from the simpler equation.

Add the equations \(x+y=6\) and \(3x-y=6\). The \(y\)-terms cancel, giving \(4x=12\), so \(x=3\). Hence, option C is correct. For example, \(x=2\) cannot satisfy both equations together. Exam tip: add two equations when a variable has equal coefficients with opposite signs.

Adding both equations gives (3x=12), so (x=4) and (y=1). Eliminate first, then find the other variable.