# Find the solution to each system of equations by the substitution method.Check answers.write you your solution in the form (x,y),(p,q) or (s,t). A)4x+5y=2 1/5x+y= -7/5

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1. `4x+5y=2` and 2. `1/5x +y=-7/5`

To use substitution convert either equation in to the form of x= or y = . This 1. becomes:

`5y = -4x +2`

`therefore y= (-4x)/5 + 2/5` which can be substituted into 2.

`1/5x + (-4x)/5 +2/5= -7/5` Rearranging the equation:

`therefore 1/5x - 4/5x = -7/5 - 2/5`

`therefore -3/5x = -9/5`

`therefore x= -9/5 divide -3/5`

`therefore x = -9/5 times -5/3` Simplify and note that the (-) symbols also cancel out

`therefore x = 3`

Now substitute into any one of your equations to find y:

`5y = -4x + 2` becomes `5y = -4(3) +2`

`5y = -12 +2`

`therefore y = - 10/5`

`therefore y = -2`

`(x;y): (3;-2)`

Now to check your answer substitute both values into one of your equations:

`4x + 5y = 2` becomes `4(3) + 5(-2) = 2`

The left hand side (LHS) must equal the right hand side(RHS)

`12 - 10 = 2`

`2=2`

Therefore LHS=RHS

(x;y):(3;-2)

`4x+5y=2`

`1/5x +y =-7/5`

take ratio of second relation away:

`x+5y= -7`

subtracting the second relation from the first:

`4x+5y -(x+5y)= 2-(-7)`

`4x+5y -x-5y= 2+7`

`3x=9` `x=3`

since:

`x+5y=-7`

`3+5y=-7`

`5y= -10` `y= -2`