Answer:
1) [tex]x=15, y=41[/tex]
2) [tex]x=6, y=24[/tex]
Step-by-step explanation:
1)
The two triangles are similar by AA similarity.
That means [tex]5x-5=4x+10[/tex].
We can add [tex]5[/tex] to both sides to get [tex]5x=4x+15[/tex].
We can then subtract [tex]4x[/tex] from both sides to get [tex]x=15[/tex].
We know that the [tex](5x-5)+(6y-4)=180[/tex] because they both lie on a straight line.
We can plug in for [tex]x[/tex] to get [tex](5(15)-5)+(6y-4)=75-5+6y-4=6y-66=180[/tex].
We can divide both sides of the equation by [tex]6[/tex] to get [tex]y-11=30[/tex].
We then add [tex]11[/tex] to both sides to get [tex]y=41[/tex].
So, [tex]\boxed{x=15, y=41}[/tex] and we're done!
2)
Because the two bases of a trapezoid are parallel, meaning adjacent top and bottom angles sum to [tex]180^{\circ}[/tex], we have that [tex]90+3y+18=180[/tex].
We can subtract a [tex]90[/tex] and an [tex]18[/tex] from both sides to get [tex]3y=72[/tex].
We then divide both sides of the equation by [tex]3[/tex] to get [tex]y=24[/tex].
We do the same thing for [tex]x[/tex]. On the other side of the trapezoid, we have that [tex]15x+30+10x=180[/tex].
We combine like terms on the left side to get [tex]25x+30=180[/tex].
We subtract a [tex]30[/tex] from both sides to get [tex]25x=150[/tex].
We divide both sides of the equation to get [tex]x=6[/tex].
So, [tex]\boxed{x=6, y=24}[/tex] and we're done!
