# 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-3y= -9 x-y=...

**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-3y= -9 x-y= -3 **

B)p+2p-4=0 7p-p-3=0

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1. `4x-3y = -9` and 2. `x-y=-3`

To use the substitution method convert 2. so that it reads x=:

`therefore x-y = -3 ` becomes `x= y-3`

Now substitute this into the other equation instead of x:

`4(y-3) - 3y = -9`

Now there is only one unknown variable (y) :

`4y - 12 -3y= -9`

`therefore 4y - 3y= -9+12`

`therefore y = 3`

Now solve for x by substituting into any one your original equations:

`therefore x=y-3` becomes `x= 3-3`

`therefore x=0`

`therefore (x;y) : (0; 3)`

Now check your answer (there are various methods) by substituting both values into any one of your equations and rendering the left-hand side(LHS) = right-hand side(RHS) :

`4x - 3y = -9` becomes `4(0) - 3(3) = -9`

`therefore 0 - 9 = -9`

`therefore -9 = -9`

**Therefore LHS=RHS**

**(x;y) : (0;3)**

**Sources:**

`a)`

`4x-3y= -9`

`x-y=-3`

multiply secon realtion by 3:

`3x-3y=-9`

subtract this one from first:

`4x-3y-(3x-3y)=-9-(-9)`

`4x-3y-3x+3y=-9+9`

`x= 0`

thus: from the second relation: `-y= -3` `y=3`

`b)`

`p+2q-4=0`

`7p-q-3 = 0`

multiply second by 2:

`14p-2q-6=0`

add to the first:

`p+2q-4+14p-2q-6=0`

`15p-10= 0` `15p= 10` `p=2/3`

From first relation we get:

`2/3 + 2q-4=0` `2q-10/3=0` `2q=10/3` `q= 5/3`