Question

solve the system of the linear equations x+2y=1,y+z=6,13z=52
​

Answer 1

Step-by-step explanation:

find z value

and put in eq. 2

Answer 2

${\boxed {x=-3,y=2,z=4}}$

${\large{\text{\underline{\underline{\red{Given :}}}}}}$

$x+2y=1$

$y+z=6$

$13z=52$

${\large{\text{\underline{\underline{\red{To Find :}}}}}}$

$x,y,z$

${\large{\text{\underline{\underline{\red{Solution :}}}}}}$

$\text{Let}$

$x+2y=1.....(1)$

$y+z=6.....(2)$

$13z=52.....(3)$

${\text{\purple{Evaluate (3)}}}$

$13z=52\implies z=\frac{52}{13}$

${\boxed {z=4}}$

${\purple{\text{Substitute the value of z in (2)}}}$

$y+z=6 \implies y+4=6$

$y=6-4$

${\boxed{y=4}}$

${\text{\purple{Substitute the value of y in (1)}}}$

$x+2y=1\implies x+2(2)=1$

$x+4=1\implies x=1-4$

${\boxed{x=-3}}$

${\large{\text{\underline{\underline{\red{Verification :}}}}}}$

$x=-3,y=2,z=4$

$\implies x+2y=1\\\implies -3+2(2)=1\\\implies-3+4=1 \\\implies1 = 1$

$\implies y+z=6\\\implies 2+4=6\\\implies 6=6$

$\implies 13z=52\\\implies 13(4)=52\\\implies 52=52$

${\pink{\textbf{Hence verified }}}$