Whenever you are given a set of linear equations and you are asked to solve for the system, you are looking to solve for the point (x, y) where the two lines intersect graphically. Algebraically, there are a couple of methods you can use to solve for such systems, those being the elimination method and the substitution method. The elimination method requires that you first eliminate one of the variables and solve for the remaining variable then use that solution to solve for the variable that was eliminated. In the substitution method, we solve one of the equations for one of the variables to yield an expression in terms of the other variable then substitute this expression into the other equation and solve for the variable involved, for which the solution is then used to solve for the other variable.

2a) x + y = 7 and 2x + 3y = 18

multiply the first equation by -2 then add it to the second equation to eliminate the x variable:

-2(x + y = 7) ==> -2(x) + -2(y) = -2(7) ==> -2x - 2y = -14

-2x - 2y = -14

+ 2x + 3y = 18

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-2x + 2x - 2y + 3y = -14 + 18 ==> 0x + 1y = 4 ==> 1y = 4 ==> y = 4

x + y = 7 and y = 4 ==> x + 4 = 7

- 4 - 4

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x + 4 - 4 = 7 - 4 ==> x = 3

solution: (3, 4)

2b) solve using the same method as 2a. notice that the x variable can be eliminated by multiplying the first equation by 3 then adding it to the second equation.

3) y = 12x + 5 and y = 7x - 10

since y is equal to 12x + 5 and 7x - 10, we can set these two expressions equal to one another. after which we can solve for x then use its solution to solve for y.

12x + 5 = 7x - 10

subtract 7x from both sides of the equation then subtract 5 from both sides of the equation. doing so yields the following:

12x - 7x + 5 - 5 = 7x - 7x - 10 - 5

5x + 0 = 0 - 15

5x = -15

divide both sides by 5: 5x/5 = -15/5 ==> x = -3

solve for y: y = 7x - 10

y = 7(-3) - 10

y = -21 - 10

y = -31

solution: (-3, -31)

4a) y = 3x + 6 and 7 + 4y = 55

this problem asks you to solve using the substitution method. since the first equation is already solved for y in terms of x, substitute it into the second equation:

7 + 4y = 55

7 + 4(3x + 6) = 55

7 + 4(3x) + 4(6) = 55

7 + 12x + 24 = 55

12x + 31 = 55

12x = 24

x = 2

y = 3x + 6

y = 3(2) + 6

y = 6 + 6

y = 12

solution: (2, 6)

4b) solving using the substitution method as was done in 4a