Question

How do you solve the following linear system: # 5x+8y=1, -2x+3y=4, #?

Answer 1

It can be solved by,
- Elimination method.
- Substitution method.

Explanation:

1)By elimination method,
`5x+8y=1--(I) eqn`
`2x+3y=4--(II)eqn`
multiply eqn (I) by 2 & eqn (II) by 5,
`10x+16y=2--(3)eqn`
`10x+15y=20--(4)eqn`
after solving further, will give
`y=-18`
so, `x=29`

2)By substitution method,
`5x+8y=1`
`x=(1-8y)/5--(1)`
`2x+3y=4--(2)`
put value of x in eqn(2)
`2((1-8y)/5)+3y=4`
`(2-16y)/5+3y=4`
`(2-16y+15y)/5=4`
`2-y=20`
`y=-18`
so, `x=(1-8(-18))/5`
`x=(1+144)/5`
`x=145/5`
`x=29`