Answer: (a) Intercepts: \((0,1)\), \(\left(-\frac12,0\right)\), \((0,5)\), and \(\left(\frac53,0\right)\); (b) \(\displaystyle x=\frac45\) or \(x=6\).

Answer: (a) Intercepts: \((0,1)\), \(\left(-\frac12,0\right)\), \((0,5)\), and \(\left(\frac53,0\right)\); (b) \(\displaystyle x=\frac45\) or \(x=6\).

(a) For \(y=\left|2x+1\right|\), the vertex is where \(2x+1=0\), so

\(x=-\frac12.\)

Therefore this graph meets the \(x\)-axis at \(\left(-\frac12,0\right)\). It meets the \(y\)-axis when \(x=0\), giving \(y=1\), so the \(y\)-intercept is \((0,1)\).

For \(y=\left|5-3x\right|\), the vertex is where \(5-3x=0\), so

\(x=\frac53.\)

Therefore this graph meets the \(x\)-axis at \(\left(\frac53,0\right)\). It meets the \(y\)-axis when \(x=0\), giving \(y=5\), so the \(y\)-intercept is \((0,5)\).

(b) To solve

\(\left|2x+1\right|=\left|5-3x\right|,\)

use the two possible cases:

\(2x+1=5-3x\)

or

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

The first case gives

\(5x=4,\)

so

\(x=\frac45.\)

The second case gives

\(2x+1=-5+3x,\)

so

\(x=6.\)