Answer:
a = 3.8
b = 1.2
Step-by-step explanation:
Given system of equations:
[tex]\begin{cases}x+y=5 \\2x-3y =4\end{cases}[/tex]
To solve the given system of equations, begin by rearranging the first equation to isolate y:
[tex]x+y=5\\\\y=5-x[/tex]
Now, substitute y = 5 - x into the second equation and solve for x:
[tex]2x-3(5-x)=4\\\\2x-15+3x=4 \\\\ 5x - 15 = 4 \\\\5x = 4 + 15 \\\\5x = 19 \\\\x=\dfrac{19}{5} \\\\ x = 3.8[/tex]
Finally, substitute x = 3.8 into y = 5 - x and solve for y:
[tex]y = 5 - 3.8 \\\\ y = 1.2[/tex]
Given that the solutions are x = a and y = b, then the values of a and b are:
[tex]\Large\boxed{\boxed{\begin{array}{l}a=3.8\\b=1.2\end{array}}}[/tex]
[tex]\dotfill[/tex]
Additional Notes
If you need the values of a and b as improper fractions, then:
[tex]a=\dfrac{19}{5}\\\\\\b=\dfrac{6}{5}[/tex]
Answer:
a = 3.8 , b = 1.2
Step-by-step explanation:
given the system of equations
x + y = 5 → (1)
2x - 3y = 4 → (2)
multiplying (1) by 3 and adding the result to (2) will eliminate y
3x + 3y = 15 → (3)
add (2) and (3) term by term to eliminate y
(2x + 3x ) + (- 3y + 3y ) = 4 + 15
5x + 0 = 19
5x = 19 ( divide both sides by 5 )
x = 3.8
substitute x = 3.8 into either of the 2 original equations and solve for y
substituting into (1)
3.8 + y = 5 ( subtract 3.8 from both sides )
y = 1.2
given x = a and y = b , then a = 3.8 and b = 1.2