Substitution

Let's say 5r-s=5 is equation 1 and -4+5s=17 is equation 2
In equation one you want to isolate s because it's easiest in this scenario.
First move the 5r to the right of the = sign so it becomes -s=5r + 5
now you need to get rid of the negative on the s so multiply everything in the equation on both sides by -1
so it becomes s=-5r-5
now substitute that into equation 2 so it becomes -4+5(-5r-5)=17
it should become -4-25r-25=17, now move the 25 and -4 over to the right side of the = sign
it should become -25r=17+25+4
after adding it should be -25r=46
now you can divide both -25r and 46 by -25 because you want to isolate r
so it becomes r=-1.84

Hi imfifteen,
That was a great answer except that you made one little error with a sign and this was carried on down the problem,
I have used your solution because it is a good one but I have made some corrections in red

Hi Brownies,
Imfifteen has isolated s in the first equation then
substituted it into the second equation.
this means that you now have an equation with only r's in it. You have to solve this for r.
lastly you substitute your r into one of the early equations to get the value for s

imfifteen:

Let's say 5r-s=5 is equation 1 and -4+5s=17 is equation 2
In equation one you want to isolate s because it's easiest in this scenario.
First move the 5r to the right of the = sign so it becomes -s=-5r + 5
now you need to get rid of the negative on the s so multiply everything in the equation on both sides by -1
so it becomes s=+5r-5 I'll call this equation 3
now substitute that into equation 2 so it becomes -4+5(+5r-5)=17
it should become -4+25r-25=17, now move the 25 and -4 over to the right side of the = sign
it should become +25r=17+25+4
after adding it should be +25r=46
now you can divide both +25r and 46 by +25 because you want to isolate r
so it becomes r=+1.84

Now substitute r=1.84 into one of the the early equations to get the value of s. (Equation 3 would be easiest because s is already the subject)