An algebra problem by Natchanon Auewetchanichkul

It was given that, $z = 25$

$3x + 1= y$ .................(1)

$2y + 5= z$ /.................(2)

Now, substitute the value of $z$ in equation (2) to find $y$

$2y + 5 = 25$

$⇒2y = 20$

$⇒y = 10$ ;[dividing 2 with 20]

Now, substitute the value of $y$ in equation (1) to find $x$

$3x + 1 = 10$

$⇒3x = 9$

$⇒x = 3$ ;[dividing 9 with 3]

Now we have all the values of $x , y , z$

Therefore , $10 \times 25 + 3$ $= 253$

put z=25 in 2y+5=z n get y=10 put y=10 in 3x+1=y to get x=3 z y=25 10=250 250+3=253

Given, $z = 25 \\ {\text{substitute this value in }} 2y + 5 = z, {\text{we get}} \Rightarrow y = 10 \\ {\text{subsituting this value in }} 3x + 1 = y, {\text{we get}} \Rightarrow x = 3 \\ So, {\color{#302B94}{answer}} = yz +x = (10)(25) + 3 = 250 + 3 = \boxed{\color{#D61F06}{253}}$