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allsm [11]
3 years ago
13

Substitue the expression 5-3x into the second equation

Mathematics
1 answer:
Rina8888 [55]3 years ago
4 0

Answer:

<h2><u>X= 1</u></h2>

Step-by-step explanation:

<em>y = 5 - 3x</em>

<em>5x - 4y = -3</em>

you substitute for y since you know what y equals.

so, it will then be

5x -4(5 -3x) = -3

you then multiply -4 by everything in the parenthesis

so then it will be

5x - 20 + 12x = -3

Then combine like terms

5x + 12x = 17x

then the new equation is

17x -20 = -3

the next step is to and 20 to both sides since it is negative you do the opposite.

17x - 20 = -3

<u>      + 20     +20</u>

17x = 17 divide 17 by both sides sice 17 is being multiplied by x you do the opposite and divide and you get 1.

<h2><u>so, X = 1</u></h2>
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Step-by-step explanation:

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Answer:

\displaystyle   \begin{cases} \displaystyle  {x} _{1} =  - p \\   \displaystyle x _{2}   =  -  q \end{cases}

Step-by-step explanation:

we would like to solve the following equation for x:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}  +  \frac{1}{x}  =  \frac{1}{p  + q + x}

to do so isolate \frac{1}{x} to right hand side and change its sign which yields:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}    =  \frac{1}{p  + q + x}  -  \frac{1}{x}

simplify Substraction:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}    =  \frac{x - (q + p +  x)}{x(p  + q + x)}

get rid of only x:

\displaystyle  \frac{1}{p}  +  \frac{1}{q}    =  \frac{  - (q + p )}{x(p  + q + x)}

simplify addition of the left hand side:

\displaystyle  \frac{q + p}{pq}     =  \frac{  - (q + p )}{x(p  + q + x)}

divide both sides by q+p Which yields:

\displaystyle  \frac{1}{pq}     =  \frac{  -1}{x(p  + q + x)}

cross multiplication:

\displaystyle    x(p  + q + x)  =   - pq

distribute:

\displaystyle    xp  + xq +  {x}^{2} =   - pq

isolate -pq to the left hand side and change its sign:

\displaystyle    xp  + xq +  {x}^{2} + pq =  0

rearrange it to standard form:

\displaystyle   {x}^{2} +    xp  + xq  + pq =  0

now notice we end up with a <u>quadratic</u><u> equation</u> therefore to solve so we can consider <u>factoring</u><u> </u><u>method</u><u> </u><u> </u>to use so

factor out x:

\displaystyle  x( {x}^{} +   p ) + xq  + pq =  0

factor out q:

\displaystyle  x( {x}^{} +   p ) +q (x + p)=  0

group:

\displaystyle  ( {x}^{} +   p ) (x + q)=  0

by <em>Zero</em><em> product</em><em> </em><em>property</em> we obtain:

\displaystyle   \begin{cases} \displaystyle  {x}^{} +   p  = 0 \\   \displaystyle x + q=  0 \end{cases}

cancel out p from the first equation and q from the second equation which yields:

\displaystyle   \begin{cases} \displaystyle  {x}^{}   =  - p \\   \displaystyle x  =  -  q \end{cases}

and we are done!

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Step-by-step explanation:

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Step-by-step explanation:

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