Split the modulus into two cases.

\(4x-3=7\) or \(4x-3=-7\).

From \(4x-3=7\), we get \(4x=10\), so \(x=\tfrac52\).

From \(4x-3=-7\), we get \(4x=-4\), so \(x=-1\).

So the solutions are \(x=-1\) and \(x=\tfrac52\).

Split into two cases.

\(1-2x=5\) or \(1-2x=-5\).

From \(1-2x=5\), we get \(-2x=4\), so \(x=-2\).

From \(1-2x=-5\), we get \(-2x=-6\), so \(x=3\).

So the solutions are \(x=-2\) and \(x=3\).

Split into two cases.

\(\dfrac{3x-2}{5}=4\) or \(\dfrac{3x-2}{5}=-4\).

From the first, \(3x-2=20\), so \(3x=22\) and \(x=\tfrac{22}{3}\).

From the second, \(3x-2=-20\), so \(3x=-18\) and \(x=-6\).

So the solutions are \(x=-6\) and \(x=\tfrac{22}{3}\).

Split into two cases.

\(\dfrac{x}{3}+2=3\) or \(\dfrac{x}{3}+2=-3\).

From the first, \(\dfrac{x}{3}=1\), so \(x=3\).

From the second, \(\dfrac{x}{3}=-5\), so \(x=-15\).

So the solutions are \(x=-15\) and \(x=3\).

Split into two cases.

\(\dfrac{x+2}{3}-\dfrac{2x}{5}=2\) or \(\dfrac{x+2}{3}-\dfrac{2x}{5}=-2\).

Multiply through by \(15\).

First case: \(5(x+2)-6x=30\), so \(-x+10=30\), hence \(x=-20\).

Second case: \(5(x+2)-6x=-30\), so \(-x+10=-30\), hence \(x=40\).

So the solutions are \(x=-20\) and \(x=40\).

Because the right-hand side equals a modulus, we must have \(3x\ge 0\), so \(x\ge 0\).

Now split into cases.

\(2x+7=3x\) gives \(x=7\).

\(2x+7=-3x\) gives \(5x=-7\), so \(x=-\tfrac75\), but this does not satisfy \(x\ge0\).

So the only solution is \(x=7\).

Use either splitting or squaring.

Method: split into two cases.

\(\dfrac{2x+1}{x-2}=5\) or \(\dfrac{2x+1}{x-2}=-5\).

First case: \(2x+1=5x-10\), so \(3x=11\) and \(x=\tfrac{11}{3}\).

Second case: \(2x+1=-5x+10\), so \(7x=9\) and \(x=\tfrac97\).

Also \(x\ne2\), and neither solution is excluded.

So the solutions are \(x=\tfrac97\) and \(x=\tfrac{11}{3}\).

Split into two cases.

\(\dfrac{3x-1}{x+5}=1\) or \(\dfrac{3x-1}{x+5}=-1\).

First case: \(3x-1=x+5\), so \(2x=6\) and \(x=3\).

Second case: \(3x-1=-x-5\), so \(4x=-4\) and \(x=-1\).

Also \(x\ne-5\), so both are valid.

So the solutions are \(x=-1\) and \(x=3\).

First simplify inside the modulus:

\(2-\dfrac{x+2}{x-3}=\dfrac{2x-6-x-2}{x-3}=\dfrac{x-8}{x-3}\).

So \(\left|\dfrac{x-8}{x-3}\right|=5\).

Then \(|x-8|=5|x-3|\).

Square both sides:

\((x-8)^2=25(x-3)^2\).

\(x^2-16x+64=25x^2-150x+225\)

\(24x^2-134x+161=0\)

\((4x-7)(6x-23)=0\)

So \(x=\tfrac74\) or \(x=\tfrac{23}{6}\).

Since the right-hand side equals a modulus, we must have \(x+1\ge0\).

Now split into cases.

\(3x-5=x+1\) gives \(2x=6\), so \(x=3\).

\(3x-5=-(x+1)\) gives \(4x=4\), so \(x=1\).

Both values satisfy \(x+1\ge0\).

So the solutions are \(x=1\) and \(x=3\).

Rearrange first:

\(|x+4|=8-x\).

Now split into two cases.

\(x+4=8-x\) or \(x+4=x-8\).

The first gives \(2x=4\), so \(x=2\).

The second gives \(4=-8\), which is impossible.

So the only solution is \(x=2\).

Rearrange first:

\(|1-2x|=8-x\).

Now split into two cases.

Case 1: \(1-2x=8-x\). Then \(-x=7\), so \(x=-7\).

Case 2: \(1-2x=-(8-x)=x-8\). Then \(-3x=-9\), so \(x=3\).

So the solutions are \(x=-7\) and \(x=3\).

Since \(|A|=|B|\), we can write \(A=B\) or \(A=-B\).

So \(2x+1=x\) or \(2x+1=-x\).

The first gives \(x=-1\).

The second gives \(3x=-1\), so \(x=-\tfrac13\).

So the solutions are \(x=-1\) and \(x=-\tfrac13\).

Since \(|A|=|B|\), solve \(A=B\) or \(A=-B\).

So \(3-2x=3x\) or \(3-2x=-3x\).

The first gives \(3=5x\), so \(x=\tfrac35\).

The second gives \(3=-x\), so \(x=-3\).

So the solutions are \(x=-3\) and \(x=\tfrac35\).

Since \(|A|=|B|\), solve \(A=B\) or \(A=-B\).

So \(2x-5=1-x\) or \(2x-5=-(1-x)\).

The first gives \(3x=6\), so \(x=2\).

The second gives \(2x-5=x-1\), so \(x=4\).

So the solutions are \(x=2\) and \(x=4\).

Since \(|A|=|B|\), solve \(A=B\) or \(A=-B\).

So \(3x+5=1+2x\) or \(3x+5=-(1+2x)\).

The first gives \(x=-4\).

The second gives \(5x=-6\), so \(x=-\tfrac65\).

So the solutions are \(x=-4\) and \(x=-\tfrac65\).

Square both sides:

\((x-5)^2=4(x+1)^2\).

\(x^2-10x+25=4x^2+8x+4\)

\(3x^2+18x-21=0\)

\(x^2+6x-7=0\)

\((x+7)(x-1)=0\)

So the solutions are \(x=-7\) and \(x=1\).

Clear the fraction first by multiplying both sides by \(2\):

\(6|2x-1|=|x-6|\).

Now square both sides:

\(36(2x-1)^2=(x-6)^2\).

\(36(4x^2-4x+1)=x^2-12x+36\)

\(144x^2-144x+36=x^2-12x+36\)

\(143x^2-132x=0\)

\(x(143x-132)=0\)

So \(x=0\) or \(x=\tfrac{132}{143}=\tfrac{12}{13}\).

Split into two cases.

\(x^2-2=7\) or \(x^2-2=-7\).

The first gives \(x^2=9\), so \(x=\pm3\).

The second gives \(x^2=-5\), which has no real solution.

So the real solutions are \(x=-3\) and \(x=3\).

Since the right-hand side must be non-negative, we need \(3-x\ge0\), so \(x\le3\).

Now split into cases.

Case 1: \(5-x^2=3-x\).

This gives \(x^2-x-2=0\), so \((x-2)(x+1)=0\).

Hence \(x=2\) or \(x=-1\).

Case 2: \(5-x^2=-(3-x)=x-3\).

This gives \(x^2+x-8=0\).

So \(x=\dfrac{-1\pm\sqrt{33}}{2}\).

All four values satisfy \(x\le3\).

So the solutions are \(-1, 2, \dfrac{-1-\sqrt{33}}{2}, \dfrac{-1+\sqrt{33}}{2}\).

Since the right-hand side must be non-negative, we need \(x+2\ge0\), so \(x\ge-2\).

Now split into cases.

Case 1: \(x^2+2x=x+2\).

This gives \(x^2+x-2=0\), so \((x+2)(x-1)=0\).

Hence \(x=-2\) or \(x=1\).

Case 2: \(x^2+2x=-(x+2)\).

This gives \(x^2+3x+2=0\), so \((x+1)(x+2)=0\).

Hence \(x=-1\) or \(x=-2\).

So the solutions are \(x=-2, -1, 1\).

Since the right-hand side must be non-negative, we need \(2x+1\ge0\), so \(x\ge-\tfrac12\).

Now split into cases.

Case 1: \(x^2-3=2x+1\).

This gives \(x^2-2x-4=0\), so \(x=1\pm\sqrt5\).

Only \(1+\sqrt5\) satisfies \(x\ge-\tfrac12\).

Case 2: \(x^2-3=-(2x+1)\).

This gives \(x^2+2x-2=0\), so \(x=-1\pm\sqrt3\).

Only \(-1+\sqrt3\) satisfies \(x\ge-\tfrac12\).

So the solutions are \(x=1+\sqrt5\) and \(x=-1+\sqrt3\).

Since the right-hand side must be non-negative, we need \(4-x\ge0\), so \(x\le4\).

Now split into cases.

Case 1: \(2x^2-5x=4-x\).

This gives \(2x^2-4x-4=0\), so \(x^2-2x-2=0\).

Hence \(x=1\pm\sqrt3\).

Case 2: \(2x^2-5x=-(4-x)=x-4\).

This gives \(2x^2-6x+4=0\), so \(x^2-3x+2=0\).

Hence \(x=1\) or \(x=2\).

All four values satisfy \(x\le4\).

So the solutions are \(x=1-\sqrt3,\ 1,\ 2,\ 1+\sqrt3\).

Since the right-hand side must be non-negative, we need \(6-x\ge0\), so \(x\le6\).

Now split into cases.

Case 1: \(x^2-7x+6=6-x\).

This gives \(x^2-6x=0\), so \(x(x-6)=0\).

Hence \(x=0\) or \(x=6\).

Case 2: \(x^2-7x+6=-(6-x)=x-6\).

This gives \(x^2-8x+12=0\), so \((x-6)(x-2)=0\).

Hence \(x=6\) or \(x=2\).

So the solutions are \(x=0,2,6\).

From \(x+2y=8\), write \(x=8-2y\).

Substitute into the second equation:

\(|(8-2y)+2|+y=6\)

\(|10-2y|+y=6\).

Now split into cases.

Case 1: \(10-2y\ge0\), so \(|10-2y|=10-2y\).

Then \(10-2y+y=6\), so \(y=4\).

Hence \(x=8-2(4)=0\).

Case 2: \(10-2y<0\), so \(|10-2y|=2y-10\).

Then \(2y-10+y=6\), so \(3y=16\) and \(y=\tfrac{16}{3}\).

Hence \(x=8-2\cdot\tfrac{16}{3}=-\tfrac83\).

So the solutions are \((0,4)\) and \(\left(-\tfrac83,\tfrac{16}{3}\right)\).

From \(3x+y=0\), write \(y=-3x\).

Substitute into \(y=|2x^2-5|\):

\(-3x=|2x^2-5|\).

Now split into cases using \(2x^2-5=\pm y\), or substitute directly into the two possibilities.

Case 1: \(2x^2-5=-3x\).

Then \(2x^2+3x-5=0\), so \((2x+5)(x-1)=0\).

This gives \(x=-\tfrac52\) or \(x=1\).

Then \(y=-3x\) gives \(y=\tfrac{15}{2}\) or \(y=-3\).

Since \(y=|2x^2-5|\ge0\), \(y=-3\) is rejected.

Case 2: \(2x^2-5=3x\).

Then \(2x^2-3x-5=0\), so \((2x-5)(x+1)=0\).

This gives \(x=\tfrac52\) or \(x=-1\).

Then \(y=-3x\) gives \(y=-\tfrac{15}{2}\) or \(y=3\).

Reject \(y=-\tfrac{15}{2}\).

So the solutions are \(\left(-\tfrac52,\tfrac{15}{2}\right)\) and \((-1,3)\).

Let \(u=|x-1|\), so \(u\ge0\).

Then the equation becomes

\(5u^2+9u-2=0\).

Factorise:

\((5u-1)(u+2)=0\).

So \(u=\tfrac15\) or \(u=-2\).

Reject \(u=-2\) because \(u\ge0\).

Thus \(|x-1|=\tfrac15\).

So \(x-1=\tfrac15\) or \(x-1=-\tfrac15\).

Hence \(x=\tfrac65\) or \(x=\tfrac45\).

Consider two cases.

Case 1: \(x\ge0\), so \(|x|=x\).

Then \(x^2-5x+6=0\), so \((x-2)(x-3)=0\).

Hence \(x=2\) or \(x=3\).

Case 2: \(x<0\), so \(|x|=-x\).

Then \(x^2+5x+6=0\), so \((x+2)(x+3)=0\).

Hence \(x=-2\) or \(x=-3\).

So the solutions are \(-3,-2,2,3\).

From solving \(x^2-5|x|+6=0\), the graph crosses the \(x\)-axis at \(x=-3,-2,2,3\).

The graph is also symmetric about the \(y\)-axis because it contains \(|x|\).

So the correct description is that it has four \(x\)-intercepts at \(-3,-2,2,3\).

The expression contains \(|x|\), so replacing \(x\) by \(-x\) leaves the function unchanged.

This means the graph is symmetric about the \(y\)-axis.

The equation of the line of symmetry is \(x=0\).

One way is to split carefully into cases, but a shorter method uses symmetry.

Testing around the turning points \(x=\pm\tfrac12\) gives the valid solutions \(x=\pm\tfrac34\).

You can also verify directly:

For \(x=\tfrac34\), \(|2x+1|+|2x-1|=\left|\tfrac52\right|+\left|\tfrac12\right|=3\).

For \(x=-\tfrac34\), \(|2x+1|+|2x-1|=\left|-\tfrac12\right|+\left|-\tfrac52\right|=3\).

So the solutions are \(x=-\tfrac34\) and \(x=\tfrac34\).

Both terms are always non-negative.

So their sum can be \(0\) only if both terms are \(0\).

Hence

\(|3x-2y-11|=0\Rightarrow 3x-2y=11\)

and

\(\sqrt{31-8x+5y}=0\Rightarrow 31-8x+5y=0\Rightarrow 8x-5y=31\).

Solve the simultaneous equations:

\(3x-2y=11\)

\(8x-5y=31\)

Multiply the first by \(5\): \(15x-10y=55\).

Multiply the second by \(2\): \(16x-10y=62\).

Subtract to get \(x=7\).

Then \(3(7)-2y=11\), so \(21-2y=11\), hence \(y=5\).

So the solution is \((x,y)=(7,5)\).