[tex]{\huge\color{yellow}{\boxed{\fcolorbox{lime}{black}{❤︎ᎯℕЅᏇᗴᖇ❦︎}}}}[/tex]

Q.1) To determine the value of [tex]x[/tex] in the equation [tex]\frac{x}{3} + \frac{5}{2} = \frac{7}{4}[/tex], we can follow the steps outlined in my previous response:

[tex]\frac{x}{3} + \frac{5}{2} = \frac{7}{4}[/tex]

To find a common denominator, we multiply the fractions:

[tex]\frac{4x}{12} + \frac{15}{12} = \frac{21}{12}[/tex]

Combine the fractions on the left-hand side:

[tex]\frac{4x + 15}{12} = \frac{21}{12}[/tex]

Multiply both sides by 12 to eliminate the denominator:

[tex]12 \times \frac{4x + 15}{12} = 12 \times \frac{21}{12}[/tex]

Simplifying further:

[tex]4x + 15 = 21[/tex]

Subtract 15 from both sides:

[tex]4x + 15 - 15 = 21 - 15[/tex]

Simplifying:

[tex]4x = 6[/tex]

Divide both sides by 4 to solve for [tex]x[/tex]:

[tex]\frac{4x}{4} = \frac{6}{4}[/tex]

[tex]x = \frac{3}{2}[/tex]

Therefore, the value of [tex]x[/tex] in the equation [tex]\frac{x}{3} + \frac{5}{2} = \frac{7}{4}[/tex] is [tex]x = \frac{3}{2}[/tex].

Q.2) To solve the system of equations:

[tex]\begin{cases} 2x + 3y = 8 \\ 4x - y = 5 \end{cases}[/tex]

There are multiple methods to solve this system, such as substitution, elimination, or matrix methods. Here, I'll use the elimination method:

Multiply the second equation by 3 to make the coefficients of [tex]y[/tex] in both equations cancel each other out:

[tex]\begin{cases} 2x + 3y = 8 \\ 12x - 3y = 15 \end{cases}[/tex]

Add the two equations together:

[tex](2x + 3y) + (12x - 3y) = 8 + 15[/tex]

Simplifying:

[tex]14x = 23[/tex]

Divide both sides by 14 to solve for [tex]x[/tex]:

[tex]\frac{14x}{14} = \frac{23}{14}[/tex]

[tex]x = \frac{23}{14}[/tex]

Now, substitute the value of [tex]x[/tex] into either of the original equations. Let's use the first equation:

[tex]2x + 3y = 8[/tex]

[tex]2\left(\frac{23}{14}\right) + 3y = 8[/tex]

Simplify and solve for [tex]y[/tex]:

[tex]\frac{46}{14} + 3y = 8[/tex]

[tex]\frac{23}{7} + 3y = 8[/tex]

Subtract [tex]\frac{23}{7}[/tex] from both sides:

[tex]3y = 8 - \frac{23}{7}[/tex]

To simplify the right-hand side, convert 8 to a fraction with a denominator of 7:

[tex]3y = \frac{56}{7} - \frac{23}{7}[/tex]

[tex]3y = \frac{33}{7}[/tex]

Divide both sides by 3 to solve for [tex]y[/tex]:

[tex]\frac{3y}{3} = \frac{33}{7} \div 3[/tex]

[tex]y = \frac{11}{7}[/tex]

Therefore, the solution to the system of equations [tex]\begin{cases} 2x + 3y = 8 \\ 4x - y = 5 \end{cases}[/tex] is [tex]x = \frac{23}{14}[/tex] and [tex]y = \frac{11}{7}[/tex].

[tex]{\huge\color{blue}{\boxed{\fcolorbox{cyan}{lavender}{✌︎TᕼᗩᑎK YOᑌ}}}}[/tex]

hope it's helps you dear ✨..

Verified answer

[tex]\huge\mathcal{\fcolorbox{orange}{azure}{\pink{➳ꪖꪀᦓ᭙ꫀ᥅ }}}[/tex]

[tex] \colorbox{pink}{\purple{sOLuTiON }}[/tex]

1) First, we can simplify the left side of the equation by multiplying the entire equation by the common denominator of 12.

This gives us [tex]4x + 30 = 21[/tex].

Then, we can isolate [tex]x[/tex] by subtracting 30 from both sides and dividing by 4.

This gives us [tex]x = -\frac{1}{6}[/tex].

2) There are many ways to solve a system of equations, but one possible method is to use substitution.

We can solve one of the equations for one of the variables, and then substitute that expression into the other equation. For example, we can solve the second equation for [tex]y[/tex] to get [tex]y = 4x - 5[/tex].

Then, we can substitute that expression into the first equation to get [tex]2x + 3(4x - 5) = 8[/tex].

Solving for [tex]x[/tex], we get [tex]x = \frac{23}{11}[/tex].

Finally, we can substitute that value back into either equation to solve for [tex]y[/tex]. Using the second equation, we get [tex]y = -\frac{2}{11}[/tex].

[tex] \colorbox{pink}{\purple{ Tнапкчѳц✨ }}[/tex]