[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]

[tex]\qquad \tt \rightarrow \:y = 1/4 [/tex]

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[tex] \large \tt Solution \: : [/tex]

[tex] \textsf{Here we go} [/tex] ~

[tex] \textsf{z varies jointly as x and y, it can be depicted as :} [/tex]

[tex]\qquad \sf  \dashrightarrow \: z \propto x \sdot \: y[/tex]

[tex] \textsf{To remove the proportionality and form} [/tex][tex] \textsf{an equation we will use a proportionality} [/tex][tex] \textsf{constant. let's say " k " } [/tex]

[tex]\qquad \sf  \dashrightarrow \: z = k(x \sdot y)[/tex]

[tex] \textsf{next we have a case when z = 120, x = 4} [/tex][tex] \textsf{and y = 5, using this information} [/tex][tex] \textsf{let's find the value of " k " } [/tex]

[tex]\qquad \sf  \dashrightarrow \: 120 = k( 4 \sdot5)[/tex]

[tex]\qquad \sf  \dashrightarrow \: 120 = k(20)[/tex]

[tex]\qquad \sf  \dashrightarrow \: k = \dfrac{120}{20} [/tex]

[tex]\qquad \sf  \dashrightarrow \: k = 6[/tex]

[tex] \textsf{Now, use this value of k, to find y,} [/tex][tex] \textsf{when x = 2, and z = 3} [/tex]

[tex]\qquad \sf  \dashrightarrow \: z = k( x \sdot y)[/tex]

[tex]\qquad \sf  \dashrightarrow \: 3 = 6(2 \sdot y)[/tex]

[tex]\qquad \sf  \dashrightarrow \: 2y = \dfrac{3}{6} [/tex]

[tex]\qquad \sf  \dashrightarrow \: y = \dfrac{1}{2} \times \dfrac{1}{2} [/tex]

[tex]\qquad \sf  \dashrightarrow \: y = \dfrac{1}{4} [/tex]