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1. In a rectangle [tex]\(PQRS\)[/tex], [tex]\(PR\)[/tex] and [tex]\(QS\)[/tex] are diagonals intersecting at [tex]\(O\)[/tex]. If [tex]\(OP = 2y + 3\)[/tex] and [tex]\(OS = 3y + 1\)[/tex], find the value of [tex]\(y\)[/tex].

Sagot :

GinnyAnswer
In a rectangle [tex]\( PQRS \)[/tex], the diagonals [tex]\( PR \)[/tex] and [tex]\( QS \)[/tex] intersect at point [tex]\( O \)[/tex]. When two diagonals of a rectangle intersect, they bisect each other. This means that the segments formed by the diagonals are equal. Therefore, we have:

[tex]\[ OP = OS \][/tex]

Given the expressions:
[tex]\[ OP = 2y + 3 \][/tex]
[tex]\[ OS = 3y + 1 \][/tex]

We can set these two expressions equal to each other since [tex]\( OP = OS \)[/tex]:

[tex]\[ 2y + 3 = 3y + 1 \][/tex]

Now, let's solve this equation for [tex]\( y \)[/tex]. First, subtract [tex]\( 2y \)[/tex] from both sides of the equation:

[tex]\[ 3 = y + 1 \][/tex]

Next, subtract 1 from both sides to isolate [tex]\( y \)[/tex]:

[tex]\[ 2 = y \][/tex]

Thus, the value of [tex]\( y \)[/tex] is:

[tex]\[ y = 2 \][/tex]
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