Question

The number of solutions of the equation.
$3x+3y-z=5$
$x+y+z=3$
$2x+2y-z=3$
(A) infinite
(B) $1$
(C) $0$
(D) two

Answer 1

The solutions of the linear equations are the values of the variable, so that when we put the value of those variables in the equation, then the equation holds for those values. A linear equation can have a unique solution, an infinite number of solutions, or two solutions. Here we have three equations and we have to find the number of solutions of the linear equations.

Complete step-by-step solution:
Linear equations define the lines in the coordinate system. There are different types of linear equations as a linear equation in one variable, a linear equation in two variables, and a linear equation in three variables.
In the above question, three equations are given to us and we have to find the number of solutions of the equation.
$3x+3y-z=5$…….eq(1)
$x+y+z=3$………eq(2)
$2x+2y-z=3$…….eq(3)
We have to find the value of x, y, and z from the above three equations.
Let us consider the first two equations. We have considered the first two equations because when we add them then the variable z will be removed from the equations as they will cancel out each other because they both are the same but having opposite signs.
On adding eq(1) and eq(2) on both sides, we get the following result.

& 3x+3y-z=5 \\\ & x\text{ }+\text{ }y\text{ }+z\text{ }=3 \\\ & \overline{4x+4y+0=\text{ }8} \\\ \end{aligned}$$ So after adding eq(1) and eq(2), we get. $$\begin{aligned} & 4x+4y=8, \\\ & \Rightarrow x+y=2 \\\ \end{aligned}$$ $$x+y=2$$…..eq(4) Now we will solve eq(1) and eq(3). On adding eq(1) and eq(3) on both sides. We get the following results. $$\begin{aligned} & 3x+3y-z=5 \\\ & 2x+2y-z=3 \\\ & \overline{5x+5y-2z=8} \\\ \end{aligned}$$ We get the equation as shown below. $$5x+5y-2z=8$$, $$\Rightarrow 5(x+y)-2z=8$$…….eq(5) On putting the value of eq(4) in eq(5), we get the following results. $$5(x+y)-2z=8$$, $$\Rightarrow 5(2)-2z=8$$, $$\begin{aligned} & \Rightarrow -2z=8-10, \\\ & \Rightarrow -2z=-2, \\\ & \Rightarrow z=1 \\\ \end{aligned}$$ So the value of z comes out to be $$1$$. From solving the above equations, we get the following results. $$x+y=2$$ $$z=1$$ If we put the value of $$x=k$$, Then the value of y is $$y=2-k$$. So from the above results, we come to know that there is an infinite number of solutions for the given system of linear equations as the lines coincide with each other. **So the correct answer is (A) infinite.** **Note:** Linear equations are also known as first order equations because the highest power of the exponent in the linear equation is one. Linear equations can be represented in many forms like the standard form, slope-intercept form, point-slope form, intercept-form, and the two-point form.