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The cost of $3$shirts and $2$sweaters is $\$114$

And the cost of$2$shirts and$4$sweaters is$\$164$

Let $x$denote the cost of each shirt.

And $y$denote the cost of each sweater.

The cost of $3$shirts and $2$sweater is $\$114$

So, $3x+2y=114$

The cost of $2$shirts and $4$sweaters is $\$164$

So,$2x+4y=164$

Solve the linar equations,

$3x+2y=114$

$2x+4y=164$

Divide the second equation by $2$,

$x+2y=82$

$x=82-2y$

Substitute $x=82-2y$in the first equation,

$3x+2y=114$

$3(82-2y)+2y=114$

$246-6y+2y=114$

$246-4y=114$

$4y=246-114$

$4y=132$

Divide both sides of the equation by $4$,

$y=33$

The cost of the sweater is$\$33$

Substitute $y=33$in the second equation,

$2x+4y=164$

$2x+4\left(33\right)=164$

$2x=164-132$

$2x=32$

Divide both sides of the equation by $2$,

$x=16$

The cost of the shirt is$\$16$

Substitute $x=16$,

$y=33$in the first equation,

$3x+2y=114$

$3\left(16\right)+2\left(33\right)=114$

$48+66=114$

$114=114$

Hence the equation holds true.