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4x+y=5 ;x+y=1 ?????      aking first-degree equations with two unknowns?

Guest Jun 10, 2015

Best Answer 

 #2
avatar+14536 
+10

4x + y  = 5


  x + y = 1      | Subtraction      =>     3x = 4   =>      x = $${\frac{{\mathtt{4}}}{{\mathtt{3}}}}$$         y  = 1 - 4/3 = $${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}$$

radix  Jun 10, 2015
 #1
avatar+92781 
+10

4x+y=5 (1)

x+y=1   (2)

(1)-(2)

3x=4

x=4/3

Sub into (2)

4/3 + y = 1

y=-1/3

Melody  Jun 10, 2015
 #2
avatar+14536 
+10
Best Answer

4x + y  = 5


  x + y = 1      | Subtraction      =>     3x = 4   =>      x = $${\frac{{\mathtt{4}}}{{\mathtt{3}}}}$$         y  = 1 - 4/3 = $${\mathtt{\,-\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}$$

radix  Jun 10, 2015
 #3
avatar+19638 
+5

4x+y=5 ;x+y=1 ?????      aking first-degree equations with two unknowns ?

 

$$\small{\text{$
\begin{array}{lccrcl}
(1)&:& & 4x + y &=& 5 \\
(2)&:& & x + y &=& 1 \\
\hline
&\\
(1) - (2)&:& & 4x - x + y - y &=& 5-1\\
& & & 3x &=& 4 \qquad |\qquad : 3\\
& & & \textcolor[rgb]{1,0,0}{x} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{ \frac{4}{3} }\\
\hline
&\\
(2)&:& & x + y &=& 1 \\
& & & \frac{4}{3} +y &=& 1 \qquad |\qquad -\frac{4}{3}\\
& & & y &=& 1 -\frac{4}{3} \\
& & & \textcolor[rgb]{1,0,0}{y} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{-\frac{1}{3}} \\
\end{array}
$}}$$

 

heureka  Jun 10, 2015

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