Question - Solving System of Equations with Two Variables
Solution:
It seems you have provided an image of two system of equations marked as 175. (a) and 176. (a) from a textbook. To solve the system in 176. (a), which is:$$\begin{cases}3(x - 1) = 4y + 1, \\5(y - 1) = x + 1,\end{cases}$$Let's solve it step by step.1. Expand the first equation:\[ 3x - 3 = 4y + 1 \]2. Now, arrange the terms to isolate x in terms of y:\[ 3x = 4y + 4 \]\[ x = \frac{4y + 4}{3} \]Now let's take this expression for x and substitute it into the second equation.3. Expand the second equation:\[ 5y - 5 = x + 1 \]Now substitute $$ x $$ with $$ \frac{4y + 4}{3} $$:\[ 5y - 5 = \frac{4y + 4}{3} + 1 \]4. Multiply through by 3 to get rid of the fraction:\[ 3(5y - 5) = 4y + 4 + 3 \]\[ 15y - 15 = 4y + 7 \]5. Subtract 4y from both sides and add 15 to both sides:\[ 15y - 4y = 7 + 15 \]\[ 11y = 22 \]6. Divide by 11 to solve for y:\[ y = \frac{22}{11} \]\[ y = 2 \]Now that we have y, we can substitute it back into the expression we derived for x:\[ x = \frac{4(2) + 4}{3} \]\[ x = \frac{8 + 4}{3} \]\[ x = \frac{12}{3} \]\[ x = 4 \]Therefore, the solution to the system is $$ x = 4 $$ and $$ y = 2 $$.
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