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3 years ago

What should be done to these equations in order to solve a system of equations by elimination?

Mathematics

2 answers:

Veronika [31]3 years ago

6 0

Answer:

multiply equn q by 3 .

when u will multiply equn 1 by 3 then you will get,9x+3y=39 and second equn is4x_3y=13 so, y can be cancelled and get the value of x for the first time and by putting the value of x you will find the value of y.

IgorC [24]3 years ago

4 0

Multiply equn q by 3.

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What is the value of x

Elenna [48]

Answer:

The value of X is 4 and the value of Y is 3

Step-by-step explanation:

4 0

3 years ago

The specifications for a new cardboard container require that the width for the container be 4 inches less than the length and t

Levart [38]

Answer:

c) Length = 12, width = 8 and height = 23.

Step-by-step explanation:

Let the length be x, then the width = x - 4 and the height = 2x - 1.

a) The polynomial is x(x - 4)(2x - 1)

b) x(x - 4)(2x - 1) = 2208

c) x(x - 4)(2x - 1) - 2280 = 0

2x^3 - 9x^2 + 4x - 2280 = 0

I'll use graphical software to find a solution for this. Its easily obtained online.

There is one real root and its x = 12.

So the dimensions of the box are L = 12, W = 8 and H = 23 inches.

6 0

2 years ago

Read 2 more answers

Which power of 2 is equal to this expression: (2•2•2)-2

zhenek [66]

2^3 is equivalent because if you count the number of twos there are in the parentheses its 3.

4 0

3 years ago

Read 2 more answers

the line joining A(a, 3) to B(2 -3) is perpendicular to the line joining C(10,1) to B. The value of a is?

RideAnS [48]

Well, first off, let's find what is the slope of BC

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
B&({{ 2}}\quad ,&{{ -3}})\quad 
% (c,d)
C&({{ 10}}\quad ,&{{ 1}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{1-(-3)}{10-2}\implies \cfrac{1+3}{10-2}
\\\\\\
\cfrac{4}{8}\implies \cfrac{1}{2}$

now, a line perpendicular to that one, will have a negative reciprocal slope, thus

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{1}{2}\\\\
slope=\cfrac{1}{{{ 2}}}\qquad negative\implies -\cfrac{1}{{{ 2}}}\qquad reciprocal\implies - \cfrac{{{ 2}}}{1}\implies \boxed{-2}$

now, we know the slope "m" of AB is -2 then, thus

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
A&({{ a}}\quad ,&{{ 3}})\quad 
% (c,d)
B&({{ 2}}\quad ,&{{ -3}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{-3-3}{2-a}=\boxed{-2}
\\\\\\
\cfrac{-6}{2-a}=-2\implies -6=-4+2a\implies -2=2a\implies \cfrac{-2}{2}=a
\\\\\\
-1=a$

6 0

2 years ago

GIVING BRAINLIEST Try the numbers 4, 5, 6, 7 in the equation 6 ⋅ x + 5 = 12 + 5 ⋅ x

Harrizon [31]

<h3>6 is a solution to equation 6x + 5 = 12 + 5x</h3>

Solution:

Given equation is:

6x + 5 = 12 + 5x

Try the numbers 4, 5, 6, 7 in the equation to test whether any of them is a solution

Substitute x = 4

6(4) + 5 = 12 + 5(4)

24 +5 = 12 + 20

$29 \neq 32$

Thus 4 is not a solution

Substitute x = 5

6(5) + 5 = 12 + 5(5)

30 + 5 = 12 + 25

$35\neq 37$

Thus 5 is not a solution

Substitute x = 6

6(6) + 5 = 12 + 5(6)

36 + 5 = 12 + 30

$41\neq 42$

Thus 6 is not a solution

Substitute x = 7

6(7) + 5 = 12 + 5(7)

42 + 5 = 12 + 35

47 = 47

Thus 6 is a solution

6 0

2 years ago