# { -0.3x + 0.1y = -0.1    6x - 2y = 2Solve each system by the addition method, if possible.

hala718 | High School Teacher | (Level 1) Educator Emeritus

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-0.3x +0.1y=-0.1 ......(1)

6x -2y =2 ............(2)

To solve the problem using the addition method, first we need to multiply one of the equation constant number in order to eliminate one variable from the equation.

If we multiply equation (1) with -20 we get:

6x -2y =2  which is equation number (2). That means that we have one equation with 2 variables. Therefore we have unlimited x and y values.

However, there is a relation between x and y values. Then the solution is:

y= 3x-1  ;where x belongs to R.

So, x could be any real number (r) and y will be (3r-1)

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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Let's try to write 0.3 and 0.1 as ratios:

0.3=3/10

0.1=1/10

Let's do the same with the coefficients of the second equation:

6=60/10

2=20/10

Now, let's re-write the equations of the system:

-3x/10 + y/10 = -1/10  (1)

60x/10 - 20y/10 = 20/10  (2)

To find the values of x and y, we have to reduce an unknown,in our case y, by multiplying the first equation by 20. After multiplying, we'll add the first equation to the second and we'll get:

-60x/10+60x/10+20y/10-20y/10=-20/10+20/10

After reducing similar terms, we'll get:

0=0

We'll conclude that the values for x and y could be any real number, as long as the pair of values respect the constraint:

6x - 2y = 2

We'll divide by 2 and we'll get:

3x-y = 1

y = 3x+1

krishna-agrawala | College Teacher | (Level 3) Valedictorian

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The given simultaneous equations are:

- 0.3x + 0.1y = - 0.1   ...   (1)

6x - 2y = 2   ...   (2)

Multiplying equation (1) by -20 we get:

6x - 2y = 2

This equation is same as equation (2). That is, the two given simultaneous equation are equivalent to each other.

Therefore, there can be no unique solution for values of x and y for these 2 equations.

neela | High School Teacher | (Level 3) Valedictorian

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To solve -0.3x+0.1y=-0.1 .........(1)and

6x-2y = 2..............................(2)

Solution:

We know that an equation could be mutplied by a constant. This action does not change the solution of the equation.

So, Eq(1) multiplied by -20 gives : 6x-2y=2 which is the eq(2). So both eq(1) and eq (2) represent the same equation.

So 6x-2y = 2  Or dividing by gives 3x-y = 1. Or

y = 3x-1 are the same equation whose solutions   are the solutions of the given 2 equations.

Since this is one equation wih 2 variables, x and y, the solutions are  many. In fact all the pairs of coordinate points, (x,y) on the graph of the straight line y = 3x-1. Some particular solutions:

x      =      0,      5,        10,       15 and the corresponding y values are:

y- 3x-1 = -1,     -14,      29,      44