Question

in exercises 1 through 10, find all solutions of the linear systems using elimination.Then check your solutions.

2.$$\begin{gathered} \left|4x+3y=2 \\ 7x+5y=3\right| \end{gathered}$$

Short answer

The solution of system of equation is $x=-1,y=2$

Step-by-step solution

 Step 1:Transforming the system

To get the solution, we will transform the value of $x,y$

$$\begin{gathered} \left|4\mathrm{x}+3\mathrm{y}=2 \\ 7\mathrm{x}+5\mathrm{y}=3\right| \end{gathered}$$ Into the form $$\begin{gathered} \left|x=.... \\ y=.... \\ \right| \end{gathered}$$

Eliminating the variables

In the given system of equations, we can eliminate the variables by adding or subtracting the equations.

In this system, we can eliminate the variable x from equation 1. First of all, multiply the first equation by 7 and multiply the second equation by 4.

$$\begin{gathered} \left|28x+21y=14 \\ 28x+20y=12\right| \end{gathered}$$

Now subtract the second equation from the first equation we get.

$$\begin{gathered} \left|28x-28x+21y-20y=14-12 \\ 28x+20y=12\right|=\left|y=2 \\ 28x+20y=12\right| \end{gathered}$$

Now put the value of y in thesecond equation.

$$\begin{gathered} \left|y=2 \\ 28x+20\left(2\right)=12\right|=\left|y=2 \\ 28x=12-40\right|=\left|y=2 \\ x=-1\right| \end{gathered}$$

Checking the solution

Now check the solution by putting the value of x and y in the given system of equation.

$$\begin{gathered} \left|28\left(-1\right)+21\left(2\right)\stackrel{?}{=14} \\ 28\left(-1\right)+20\left(2\right)\stackrel{?}{=12}\right|\Rightarrow \left|-28+42\stackrel{?}{=14} \\ -28+40\stackrel{?}{=}12\right| \\ \Rightarrow \left|14=14 \\ 12=12\right| \end{gathered}$$

Verified.

Hence, the solution of system of equation is $x=-1,y=2$