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(B) The vertices of a triangle are the intersection points of the lines forming its sides. We solve the equations in pairs to find these points.Pair $

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$(1, 0), (3, -1), (2, 5)$

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(B) The vertices of a triangle are the intersection points of the lines forming its sides. We solve the equations in pairs to find these points.Pair $1$: $5x - y = 5$ and $x + 2y = 1$.From the first equation,$y = 5x - 5$. Substituting into the second: $x + 2(5x - 5) = 1 \implies x + 10x - 10 = 1 \implies 11x = 11 \implies x = 1$. Then $y = 5(1) - 5 = 0$. Vertex: $(1, 0)$.Pair $2$: $x + 2y = 1$ and $6x + y = 17$.From the second equation,$y = 17 - 6x$. Substituting into the first: $x + 2(17 - 6x) = 1 \implies x + 34 - 12x = 1 \implies -11x = -33 \implies x = 3$. Then $y = 17 - 6(3) = -1$. Vertex: $(3, -1)$.Pair $3$: $5x - y = 5$ and $6x + y = 17$.Adding the two equations: $(5x - y) + (6x + y) = 5 + 17 \implies 11x = 22 \implies x = 2$. Substituting $x = 2$ into $5x - y = 5$: $5(2) - y = 5 \implies 10 - y = 5 \implies y = 5$. Vertex: $(2, 5)$.The vertices are $(1, 0), (3, -1)$,and $(2, 5)$.

Part $I$	Part $II$
$1.$ The solution of $2x - y = 4$ and $3x - 2y = 5$	$a.$ $x = 3, y = 2$
$2.$ The solution of $5x - y = 14$ and $4x - 3y = 9$	$b.$ $x = 3, y = 1$
$3.$ The solution of $4x - 3y = 5$ and $3x - y = 5$	$c.$ $x = 2, y = 1$
$4.$ The solution of $3x - 2y = -1$ and $2x - 5y = -8$	$d.$ $x = 1, y = 2$

Determine,algebraically,the vertices of the triangle formed by the lines $5x - y = 5$,$x + 2y = 1$,and $6x + y = 17$.

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