GKT101: General Knowledge for Teachers – Math

1.

The $y$-intercept is the point where the graph intersects the $y$-axis. Since the $y$-axis is also the line $x = 0$, the $x$-value of this point will always be $0$.

The $x$-intercept is the point where the graph intersects the $x$-axis. Since the $x$-axis is also the line $y = 0$, the $y$-value of this point will always be $0$.

To find the $y$-intercept, let's substitute $x = 0$ into the equation and solve for $y$:

$\begin{aligned} 4 \cdot 0-3 y &=17 \\ -3 y &=17 \\ y &=-\frac{17}{3} \end{aligned}$

So the $y$-intercept is $\left(0,-\frac{17}{3}\right)$.

To find the $x$ intercept, let's substitute $y = 0$ into the equation and solve for $x$:

$\begin{array}{r} 4 x-3 \cdot 0=17 \\ 4 x=17 \\ x=\frac{17}{4} \end{array}$

So the $x$-intercept is $\left(\frac{17}{4}, 0\right)$.

In conclusion,

• The $y$-intercept is $\left(0,-\frac{17}{3}\right)$.
• The $x$-intercept is $\left(\frac{17}{4}, 0\right)$.

2.

The $y$-intercept is the point where the graph intersects the $y$-axis. Since the $y$-axis is also the line $x = 0$, the $x$-value of this point will always be $0$.

The $x$-intercept is the point where the graph intersects the $x$-axis. Since the $x$-axis is also the line $y = 0$, the $y$-value of this point will always be $0$.

To find the $y$-intercept, let's substitute $x = 0$ into the equation and solve for $y$:

$\begin{aligned} y-3 &=5(0-2) \\ y-3 &=-10 \\ y &=-7 \end{aligned}$

So the $y$-intercept is $(0, -7)$.

To find the $x$ intercept, let's substitute $y = 0$ into the equation and solve for $x$:

$\begin{aligned} 0-3 &=5(x-2) \\ -3 &=5 x-10 \\ 7 &=5 x \\ 1.4 &=x \end{aligned}$

So the $x$-intercept is $(1.4, 0)$.

In conclusion,

• The $y$-intercept is $(0, -7)$.
• The $x$-intercept is $(1.4, 0)$.

3.

The $x$-intercept is the point where the graph intersects the $x$-axis. Since the $x$-axis is also the line $y = 0$, the $y$-value of this point will always be $0$.

The $y$-intercept is the point where the graph intersects the $y$-axis. Since the $y$-axis is also the line $x = 0$, the $x$-value of this point will always be $0$.

To find the $x$ intercept, let's substitute $y = 0$ into the equation and solve for $x$:

$\begin{aligned} 0 &=11 x+6 \\ -6 &=11 x \\ -\frac{6}{11} &=x \end{aligned}$

So the $x$-intercept is $\left(-\frac{6}{11}, 0\right)$.

To find the $y$-intercept, let's substitute $x = 0$ into the equation and solve for $y$:

$\begin{aligned} &y=11 \cdot 0+6 \\ &y=6 \end{aligned}$

So the $y$-intercept is $(0, 6)$. Generally, in linear equations of the form $y=m x+b$ (which is called slope-intercept form), the $y$-intercept is $( 0, b)$.

In conclusion,

• The $x$-intercept is $\left(-\frac{6}{11}, 0\right)$.
• The $y$-intercept is $(0, 6)$.

4.

The $y$-intercept is the point where the graph intersects the $y$-axis. Since the $y$-axis is also the line $x = 0$, the $x$-value of this point will always be $0$.

The $x$-intercept is the point where the graph intersects the $x$-axis. Since the $x$-axis is also the line $y = 0$, the $y$-value of this point will always be $0$.

To find the $y$-intercept, let's substitute $x = 0$ into the equation and solve for $y$:

$\begin{aligned} -7 \cdot 0-6 y &=-15 \\ -6 y &=-15 \\ y &=\frac{15}{6} \\ y &=\frac{5}{2} \end{aligned}$

So the $y$-intercept is $\left(0, \frac{5}{2}\right)$.

To find the $x$ intercept, let's substitute $y = 0$ into the equation and solve for $x$:

$\begin{aligned} -7 x-6 \cdot 0 &=-15 \\ -7 x &=-15 \\ x &=\frac{15}{7} \end{aligned}$

So the $x$-intercept is $\left(\frac{15}{7}, 0\right)$.

In conclusion,

• The $y$-intercept is $\left(0, \frac{5}{2}\right)$.
• The $x$-intercept is $\left(\frac{15}{7}, 0\right)$.