Question

(06 0zr mc)
what equation is solved by the graphed systems of equations?
(7, 25)
$\boldsymbol{3x + 4 = 5x - 10}$
$\boldsymbol{3x - 4 = 5x - 10}$
$\boldsymbol{3x + 4 = 5x + 10}$
$\boldsymbol{3x - 4 = 5x + 10}$

Explanation:

Step1: Recall the point of intersection

The solution to a system of linear equations (graphed as lines) is the point where they intersect. Here, the intersection point is \((7, 25)\), so we can substitute \(x = 7\) into each equation to see which one satisfies \(y = 25\) for both sides.

Step2: Test the first option \(3x + 4 = 5x - 10\)

Left - hand side (LHS) when \(x = 7\): \(3(7)+4=21 + 4=25\)
Right - hand side (RHS) when \(x = 7\): \(5(7)-10=35 - 10 = 25\)
Since LHS = RHS for \(x = 7\), we can also check the slopes and y - intercepts. The general form of a line is \(y=mx + b\), where \(m\) is the slope and \(b\) is the y - intercept.
For the line \(y = 3x+4\), slope \(m_1 = 3\), y - intercept \(b_1 = 4\).
For the line \(y=5x - 10\), slope \(m_2 = 5\), y - intercept \(b_2=- 10\).
We can also think about the slope calculation. If we consider two points on a line, slope \(m=\frac{y_2 - y_1}{x_2 - x_1}\). But since we know the intersection point and the form of the equations, substituting \(x = 7\) is a quick way.

Let's check the other options for completeness:

• For \(3x-4 = 5x - 10\), LHS when \(x = 7\): \(3(7)-4=21 - 4 = 17

eq25\)

• For \(3x + 4=5x + 10\), RHS when \(x = 7\): \(5(7)+10=35 + 10=45

eq25\)

• For \(3x-4 = 5x + 10\), LHS when \(x = 7\): \(3(7)-4 = 17\), RHS when \(x = 7\): \(5(7)+10 = 45\), \(17

eq45\)

Answer:

\(3x + 4 = 5x - 10\) (the first option)