Algebra 2 system of equations help?

-r + t = 5

-2r + t = 4

I'm going to solve this by the elimination (though substitution way is more easier), this process involves eliminating a variable of both equations by adding/subtracting the equations. In order to do that, we must have the same number or the same number but with different signs.

So, I'm going to eliminate t, since they are the same, by subtracting the equations

-r + t = 5

-2r + t = 4

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r = 1

So we got r, now to solve for t, we must substitute this r value into a r variable of one equation and solve for t.

-r + t = 5

-1 + t = 5

-1 + t + 1 = 5 + 1

t = 6

The solution to this problem is (1, 6)

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CHECK:

-1 + 6 = 5

5 = 5

-2 (1) + 6 = 4

-2 + 6 = 4

4 = 4 YES

-r + t = 5

-2r + t = 4

r = 1

t = 6

Easy. Make the first equation equal to "t". So by adding r on both sides, you get t = r + 5. Now that you have a value of "t", substitute it into the second equation, so you will get -2r + (r + 5) = 4. Expand, so you get -2r + r +5 = 4, then bring the variable to one side and the numbers to the other, so you get -r = -1, so r = 1. Then sub 1 for r in either equation (I recommend the first) and solve for t, which is 6.

-r + t = 5 .....(1)

-2r + t = 4....(2)

subtract (2) from (1)

-r + t -(-2r + t) = 5 - 4

-r + t +2r - t = 1

r = 1

plug in value of r in (1)

-(1) + t = 5

t = 5+1

t = 6