1.6 Linear Equations (HL)

Topic	1.6 Linear Equations (HL)
Tag	n/a
Source	N16-TZ0-P1-1(HL)
Question Text	$2x + 2y + 2z = 3, x - y + z = 5$ and $x + y + 4z = 6$ can be written as $(a, b, c)$ . Find $a + b + c$ .
Total Mark	5
Correct Answer	-1
Explanation	n/a
Mark Scheme	$2x + 2y + 2z = 3$ (A) $x - y + z = 5$ (B) $x + y + 2z = 6$ (C) Step 1: Eliminate one variable to form two equations with two unknown values. For this question, we will eliminate y. $(A) + 2(B)$ $4x + 4z = 8$ $x + z = 2$ (D) $(B) + (C)$ $2x + 3z = 11$ (E) Using equations $D$ and $E$ , we can find the value of $x$ and $z$ . $x = 6, z = -4$ Step 2: Find the value of the eliminated variable from step 1. $y =- 3$ Hence, the point of intersection has coordinates $(6, 3, 4)$ . Thus, $6-3-4=-1$ .

Topic	1.6 Linear Equations (HL)
Tag	n/a
Source	M16-TZ2-P1-1(HL)
Question Text	The following system of equations represents three planes in space. $x - y + z = -2$ $2x - y - 2z = -9$ $3x + y - z = -2$ The coordinates of the point of intersection of the three planes can be written as $(x, y, z)$ . Find $x + y + z$ .
Total Mark	6
Correct Answer	4
Explanation	n/a
Mark Scheme	$x - y + z = -2$ (A) $2x - y - 2z = -9$ (B) $3x + y - z = -2$ (C) Step 1: Eliminate one variable to form two equations with two unknown values. For this question, we will eliminate y. $(A) - (B)$ : $-x + 3z = 7$ (D) $(B) + (C)$ : $5x - 3z = -11$ (E) Using equations $D$ and $E$ , we can find the value of $y$ and $z$ . $x = -1,z = 2$ Step 2: Find the value of the eliminated variable from step 1. $y = 3$ Hence, the point of intersection has coordinates $(-1, 3, 2)$ . $x + y + z = 4$

Topic	1.6 Linear Equations (HL)
Tag	n/a
Source	N18/5/MATHL/HP1/ENG/TZ0/XX/4
Question Text	Consider the following system of equations where $a ∈ R$ . R1: $3x + 6y - z = 15$ R2: $x + 2y + az = 5$ R3: $5x + 12y = 2a$ The solution of the system of equations when $a = 2$ is $(x, y, z)$ . Find $x + y + z$ .
Total Mark	4
Correct Answer	-11
Explanation	n/a
Mark Scheme	When $a=2$ , we can rearrange the system of equations as below. R1: $3x + 6y - z = 15$ R2: $x + 2y + 2z = 5$ R3: $5x + 12y = 4$ Next, use elimination or row reduction to find the value of $x$ , $y$ and $z$ . $x= - 27, y = 16, z = 0$ Thus, $x + y + z = -11$