You can use the fact that when coefficient of x variable are of equal magnitude but opposite sign, then adding them will make the coefficient 0, thus, making the x variable eliminated.

The numbers that can be multiplied by each equation so that when the two equations are added together, the x term is eliminated is given by

Option A: –10 times the first equation and 3 times the second equation

<h3>How does elimination works?</h3>

This method is actually called method of elimination to solve a system of linear equations.

We make one specific variable's coefficients of equal magnitude so that we can subtract or add the equations and eliminate that variable to make it easy to get the value of the other variable which will then help in getting the value of the first variable (if working in dual variable system).

If we have equations:

$a_1x + b_1y = c_1\\a_2x + b_2y = c_2$

then, if we want to eliminate variable x, then we have to multiply equation 1 with

$- \dfrac{a_2}{a_1}$

which will make coefficient of x in first equation as

$a_1 \times - \dfrac{a_2}{a_1} = -a_1$

Then adding both equation will eliminate the variable x.

We could've skipped that -ve sign and at then end, instead of adding, we could've subtracted the equations.

<h3>What is magnitude and sign?</h3>

5 has 5 as magnitude, and sign isn't present which means its of positive (+) sign.

-5 has 5 as magnitude and sign is negative(-).

For this case, we're multiplying both the equations but the core concept or aim is same, ie, making the coefficients of equal magnitude but with opposite sign.

<h3>Using the above facts to get the numbers to multiply the equations of the given system</h3>

The given system of equations is

$\dfrac{1}{5}x + \dfrac{3}{4}y = 9\\\\\dfrac{2}{3}x - \dfrac{5}{6}y = 8$

Let two numbers be p,and q who multiply equation first and second respectively to make coefficient of x of equal magnitude but opposite sign.

We have

$\text{Coefficient of x in first equation}= a_1 = \dfrac{1}{5}\\\\\text{Coefficient of x in second equation} = a_2 = \dfrac{2}{3}$

Multiplying with p and q will give us

$p\times a_1 = \dfrac{p}{5}\\\\q \times a_2 = \dfrac{2q}{3}$

We need both resultant coefficient to add up to 0, or

$p/5 + 2q/3 = 0\\p = -10q/3\\q = -3p/10$

Now in options, we see first equation is either getting multiplied with -10, 10, or -3,3

If we put q = 3, we get p = -10q/3= -10

If we put q = -3, we get p = 10

If we put q = 5, we get p = -50/3

Thus only first choice is matching the correct pairs.

Thus,

The numbers that can be multiplied by each equation so that when the two equations are added together, the x term is eliminated is given by

Option A: –10 times the first equation and 3 times the second equation

Learn more about method of elimination here:

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7 0

2 years ago

Read 2 more answers

I am a mutiple of 60. I am between 400 and 600 . I am 40 away from a multiple of 100. who am i

crimeas [40]

There is no possible answer.
The #s are 420, 480, and 540.
None of them are 40 away from a multiple of 100.
However, it is was between 300 and 600, the answer would be 360.

3 0

4 years ago

Find the area of the polygon

Evgesh-ka [11]

Answer:

Step-by-step explanation:

The area of a parallelogram is just b*h

7 0

3 years ago