Question

Examine this system of equations. Which numbers can be multiplied by each equation so that when the two equations are added together, the y term is eliminated? One-fourth x minus one-sixth y = 5 Four-fifths x StartFraction 3 Over 8 y EndFraction = 10 18 times the first equation and 8 times the second equation –18 times the first equation and 8 times the second equation 8 times the first equation and 6 times the second equation –8 times the first equation and 6 times the second equation.

Answer 1

Answer:

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Step-by-step explanation:

Answer 2

In the elimination method one variable is eliminated from the system of equation to find the value of other.

The first equation must be multiplied by 18 and second equation must be multiplied by 8.

What is elimination method?

To solve the system of equation and find the value of the variables, the elimination method is used.

In this method one variable is eliminated from the system of equation to find the value of other.

Given information-

The system of equation given in the problem is,

$\dfrac{1}{4}x -\dfrac{1}{6}y =5$              ......1

Let the above equation as equation 1.

The second equation given in the problem is,

$\dfrac{4}{5}x +\dfrac{3}{8}y =10$             ......2

Let the above equation as equation 2.

To eliminate the y term, make the coefficient of y of both the equation equal.

As the coefficient of y of  the equation 2 is 3/8.  Thus multiply second equation with 8 to eliminate coefficient. Thus,

$8\times \dfrac{4}{5}x +8\times\dfrac{3}{8}y =8\times10$

$\dfrac{32}{5}x +3y =80$              .....3

Multiply  equation 1 with 18 to make the coefficient of y equal. Thus,

$18 \times\dfrac{1}{4}x -18\times\dfrac{1}{6} \times y =18 \times5$

$\dfrac{18}{4}x -3 =90$            ......4

Add equation 3 and equation 4 as,

$\dfrac{32}{5}x +3y+ \dfrac{18}{4}x -3y =80+90$

$\dfrac{32}{5}x + \dfrac{18}{4}x =170$

Hence the first equation must be multiplied by 18 and second equation must be multiplied by 8.

Learn more about the elimination method here;

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