Answer :
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]
[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]
