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Harman [31]
2 years ago
15

Examine this system of equations. Which numbers can be multiplied by each equation so that when the two equations are added toge

ther, the x term is eliminated? One-fifth x three-fourths y = 9 Two-thirds x minus five-sixths y = 8 –10 times the first equation and 3 times the second equation 10 times the first equation and 3 times the second equation –3 times the first equation and 5 times the second equation 3 times the first equation and 5 times the second equation.
Mathematics
2 answers:
Natali5045456 [20]2 years ago
8 0

<em>I know i am 2 weeks late but the answer is B, Have a good day :))</em>

WARRIOR [948]2 years ago
7 0

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:

brainly.com/question/20385690

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9. x=22

10. x=12

Step-by-step explanation:

1. RS/WS=ST/SX and m<RST=m<WSX

2. AB/PQ=8/6=4/3

BC/QR=AC/PR=12/9=4/3

AB/PQ=BC/QR=AC/PR

3. ST/JP=10/15=2/3

SU/JM=14/21=2/3

ST/JP=2/3=SU/JM

and m<TSU=70°=m<PJM

4. DK/PR=8/4=2

JK/ZR=18/9=2

DK/PR=2=JK/ZR

and m<DKJ=65°=m<PRZ

5. RT/ST=UT/LT

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6. KL/YW=20/18=10/9

JL/XW=36/24=3/2

KL/YW=10/9≠3/2=JL/XW

7. BF/NL=24/16=3/2

BE/NM=39/26=3/2

BF/NL=3/2=BE/NM

and m<EBF=m<MNL

8. GH/QR=32/20=8/5

HI/RS=40/25=8/5

GI/QS=24/15=8/5

GH/QR=HI/RS=GI/QS=8/5

9. x/33=18/27

Simplifying the fraction on the right side of the equation:

x/33=2/3

Solving for x: Multiplying both sides of the equation by 33:

33(x/33)=33(2/3)

x=11(2)

x=22

10. x/16=9/12

Simplifying the fraction on the right side of the equation:

x/16=3/4

Solving for x: Multiplying both sides of the equation by 16:

16(x/16)=16(3/4)

x=4(3)

x=12

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