how many solutions does this linear system have?\ny = \\frac{2}{3}x + 2\n6x - 4y = -10\none solution: (-0.6…

Explanation:

Step1: Rewrite the first - equation in standard form

The first equation $y=\frac{2}{3}x + 2$ can be rewritten as $- \frac{2}{3}x+y=2$. Multiply through by 3 to get $-2x + 3y=6$.

Step2: Rewrite the second - equation in slope - intercept form

The second equation is $6x-4y=-10$. Solve for $y$: [ \begin{align*} -4y&=-6x - 10\ y&=\frac{6}{4}x+\frac{10}{4}\ y&=\frac{3}{2}x+\frac{5}{2} \end{align*} ] The first equation in slope - intercept form is $y=\frac{2}{3}x + 2$, and the second is $y=\frac{3}{2}x+\frac{5}{2}$. The slopes of the two lines are $\frac{2}{3}$ and $\frac{3}{2}$ respectively. Since the slopes are different, the two lines intersect at exactly one point. We can also use the method of substitution. Substitute $y=\frac{2}{3}x + 2$ into $6x-4y=-10$: [ \begin{align*} 6x-4(\frac{2}{3}x + 2)&=-10\ 6x-\frac{8}{3}x-8&=-10\ \frac{18x-8x}{3}&=-10 + 8\ \frac{10x}{3}&=-2\ x&=-0.6 \end{align*} ] Substitute $x = - 0.6$ into $y=\frac{2}{3}x+2$: [ \begin{align*} y&=\frac{2}{3}\times(-0.6)+2\ y&=-0.4 + 2\ y&=1.6 \end{align*} ]

Answer:

one solution: $(-0.6,1.6)$