What strategies can you use to solve a linear equation such as?

Answer 1

linear equations are of the following forms:
ax + by = c; where a, b, c are constants
ax = c (a special case - a vertical line)
bx = c (another special case - a horizontal line)

the line you get by graphing the equation is the set of all numbers that meet the conditions of the equation

if you put the equation into slope-intercept form, y = mx + b
then m = the slope and b = the y-intercept

A related problem is how to solve for a pair of equations with two unknowns.

equation 1: 3x + 5y = 2
equation 2: 5x - 3y = -8

First solve for x in the first equation
3x = 2 - 5y
x = 2/3 -5y/3
then substitute that for x in the second equation
5(2/3 - 5y/3) - 3y = -8
10/3 - 25y/3 - 3y = -8
8 + 10/3 = 3y + 25y/3
24 + 10 = 9y + 25y
34 = 34y
1 = y
substitute for y in the first equation
3x + 5(1) = 2
3x + 5 = 2
3x = -3
x = -1
Now put x and y into the second equation to check
5x - 3y = -8
5(-1) - 3(1) = -8
-5 -3 = -8
-8 = -8
checks

That's the substitution method. There's also the addition method.

equation 1: 3x + 5y = 2
equation 2: 5x - 3y = -8

Multiply the first equation by 3 and the second equation by 5, then add them.
9x + 15y = 6
25x - 15y = -40
34x = -34
x = -1
then proceed as before

Answer 2

4 +5x = x-8. I continuously write down the equation first. 4 -4 +5x -x = x-x -8 -4. Now i visit attempt to get each and every and each of the x term on one section, and each and each of the constants or numerals on the countless. you spot, as promptly as I subtract a 4 from one section, I do only an identical on the countless. To be fairly uncomplicated, inspite of operation you do on one section, you do the precise same on the countless. 4x = -12 having divided by using way of four on the two sides, we've x = -3 only remember, it incredibly is an equation. so which you shelter it equivalent by using way of appearing the precise same ingredient on the two sides. and you attempt to isolate the variable to be solved on one section.