graph your equations: 2x + 3y = 7 and y=2x + 27 label them properly, including point of intersection?

y = -2x/3+7/3
y = 2x+27

2x+2x/3 = 7/3-27 
8x/3 = -74/3
8x = -74
x = -37/4
y = -37/2+27 = 
intersection coordinates -37/4 ; 17/2

2x+3y=7
| y | 1 | -1 |
| x | 2 | +5|

y=2x+27
| y |..7.. |..5..|
| x |-10..|-11.|
Each equation can be graphed on a graph paper
with the suggested 2 points.
Set up a rectangular coordinates system properly
on a piece of graph paper
(1) Mark the 2 points for each equation & draw the
straight line passing through them. You should get
2 lines.
(2) Mark the point of intersection of the 2 lines on
the paper, which should be at
x=-9.25
y=8.5
End.

 2x + 3y = 7 eq1 and y=2x + 27 eq2

plug in eq2 to eq1

2x + 3(2x + 27) = 7
2x + 6x + 81 = 7
8x = 7 - 81
8x = -74
 x = -37/4 Answer//


solving for y at eq2

y = 2(-37/4) + 27 = 17/2 Answer//

2x + 3y = 7, y = 2x + 27
{y = 7/3 - (2 x)/3, y = 2 x + 27}
x = -37/4, y = 17/2

As simultaneous equations solutions are x = -9.5 and y = 8.5
Therefore point of intersection is at (-9.5, 8.5) 

The typical equation of a line is: y = mx + y₀ → where m: slope and where y₀: y-intercept


2x + 3y = 7 ← this is the line (ℓ₁)

3y = - 2x + 7

y = - (2/3).x + (7/3) → the slope is (- 2/3) and the y-intercept is (7/3)


y = 2x + 27 ← this is the line (ℓ₂)

y = 2x + 27 → the slope is (2) and the y-intercept is (27)


Intersection between the line (ℓ₁) and the line (ℓ₂) → (ℓ₁) = (ℓ₂)

y = y

- (2/3).x + (7/3) = 2x + 27

(- 2x + 7)/3 = 2x + 27

- 2x + 7 = 6x + 81

- 8x = 74

x = - 74/8

x = - 37/4 ← this is the abscissa of te poiint of intersection


Recall:

y = 2x + 27 → where: x = - 37/4

y = 2.(- 37/4) + 27

y = - (37/2) + (54/2)

y = 17/2 ← this is the ordinate of te point of intersection


M (- 37/4 ; 17/2)



To draw a line, you need only 2 distinct points

2x + 3y = 7 ← this is the line (ℓ₁)

2x + 3y = 7 → when: x = 0

3y = 7

y = 7/3 → first point A (0 ; 7/3)

2x + 3y = 7 ← this is the line (ℓ₁)

2x + 3y = 7 → when: y = 0

2x = 7

x = 7/2 → second point B (7/2 ; 0)


y = 2x + 27 ← this is the line (ℓ₂)

y = 2x + 27 → when: x = 0

y = 27 → first point C (0 ; 27)

y = 2x + 27 ← this is the line (ℓ₂)

y = 2x + 27 → when: y = 0

0 = 2x + 27

2x = - 27

x = - 27/2 → second point D (- 27/2 ; 0)

The answer is 12

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